## What Is the Isosceles Triangle Theorem?

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What is the Isosceles Triangle Theorem, and why is it something you need to know? We answer this question and more.

Before we cover the Isosceles Triangle Theorem, we’ll discuss how we have used triangles over time in architecture, art, and design. Then we’ll talk about the history of isosceles triangles, the different types of triangles, and the different parts of isosceles triangles.

And lastly, after discussing the theorem, we’ll go over some useful formulas for calculating various parts of isosceles triangles.

## Various Applications of Triangles

Image via: Flickr

While rectangles are more prevalent in architecture because they are easy to stack and organize, triangles provide more strength. With modern technology, triangles are easier to incorporate into building designs and are becoming more prevalent as a result.

In 1989, Japanese architects decided that a triangular building design would be necessary if they were to construct a 500-story building in Tokyo. The triangular shape could withstand earthquake forces, unlike a rectangular or square design.

You can also see triangular building designs in Norway, the Flatiron Building in New York, public buildings and colleges, and modern home designs. ​​​​

As far as isosceles triangles, you see them in architecture, from ancient to modern. Ancient Egyptians used them to create pyramids. Ancient Greeks used obtuse isosceles triangles as the shapes of gables and pediments. In the Middle Ages, architects used what is called the Egyptian isosceles triangle, or an acute isosceles triangle.

You can also see isosceles triangles in the work of artists and designers going back to the Neolithic era. They are visible on flags, heraldry, and in religious symbols.

## The History of Isosceles Triangles

The study of triangles is almost as old as civilization. Ancient Egyptians studied them as did Babylonians. You can see how ancient Egyptians used triangles to construct pyramids.

It wasn’t until about 300 BCE that the Greek mathematician, Euclid, gave triangles with two equal sides a name. He combined the Greek words “isos” (equal) and “skelos” (legs) to define them as triangles with exactly two sides.

Today, mathematicians define isosceles triangles as having at least two equal sides. This is a subtle but important difference because it means that equilateral triangles are also considered to be isosceles triangles. More on that below.

### Other Types of Triangles

In the world of geometry, there are many types of triangles besides isosceles:

• Right triangles are triangles that have one right angle equaling 90 degrees.

• Scalene triangles are triangles with no equal sides.

• Acute triangles are triangles where all three angles are less than 90 degrees.

• Obtuse triangles have one angle that is greater than 90 degrees.

• Equilateral triangles are triangles with three equal sides and angles.

Isosceles triangles can also be acute, obtuse, or right, depending on their angle measurements. Equilateral triangles can be a type of isosceles triangle. However, note that not all isosceles triangles are equilateral. Moreover, an isosceles triangle can never be a scalene triangle.

There are a few particular types of isosceles triangles worth noting, such as the isosceles right triangle, or a 45-45-90 triangle. There is also the Calabi triangle, an obtuse isosceles triangle in which there are three different placements for the largest square.

And last but not least, there is also the golden triangle, which is an isosceles triangle where the duplicated leg is in the golden ratio to the distinct side. The golden ratio is defined as a ratio of two numbers in which the ratio of the sum to the bigger number is the same as the ratio of the larger number to the smaller.

## The Parts of Isosceles Triangles

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No matter how you define isosceles triangles, they are all made up of two legs and a base. If it’s an equilateral triangle, all sides can be considered the base because all sides are equal.

And, there are two equal angles opposite the equal sides. Each of these angles is called a base angle. The angle in between the legs is the vertex angle.

Also, the angles opposite each leg are equal and always less than 90 degrees (acute). All total, the angles should add up to 180 degrees.

## The Isosceles Triangle Theorem

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Finally, it’s time to discuss the Isosceles Triangle Theorem. The Isosceles Triangle Theorem states: In a triangle, angles the opposite to the equal sides are equal.

So, how do we go about proving it true? It’s pretty simple. First, we’re going to need to label the different parts of an isosceles triangle.

Let’s give the points of the isosceles triangle the labels A, B, and D (counterclockwise from the top). We also need to draw a line from the center of the base (BD) to the angle (A) on the other side. Note that the center of the base is termed midpoint, and angles on the inside of the triangle are called interior angles.

Where that line intersects the side is labeled C. The line creates two triangles, ABC and ACD.

So, Point C is on the base BD, creating line segment AC. Here’s what we have so far:

• BC is congruent to CD (median)

• AC is congruent to AC (reflexive property)

• AB is congruent to AD (given)

We have what is called the Side Side Postulate because all of the sides of ABC (three total) are congruent with ACD. If triangle ABC and triangle ACD have congruence, then their matching parts are congruent. This also proves that the B angle is congruent with the D angle.

## The Converse of the Isosceles Triangle Theorem

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As with most mathematical theorems, there is a reverse of the Isosceles Triangle Theorem (usually referred to as the converse). It states, “if two angles of a triangle are congruent, the sides opposite to these angles are congruent.” Let’s work through it.

First, we’ll need another isosceles triangle, EFH. The EFH angle is congruent with the EHF angle. We need to prove that EF is congruent with EH. To do that, draw a line from FEH (E is the apex angle) to the base FH. Label this point on the base as G.

By doing this, we have made two right triangles, EFG and EGH. Because we have an angle bisector with the line segment EG, FEG is congruent with HEG. So what is the result?

We now have what’s known as the Angle Angle Side Theorem, or AAS Theorem, which states that two triangles are equal if two sides and the angle between them are equal.

Let’s take a look. We know that EFG is congruent with EHF. FEG is congruent with HEG. And EG is congruent with EG. That gives us two angles and a side, which is the AAS theorem.

When the triangles are proven to be congruent, the parts of the triangles are also congruent making EF congruent with EH. By working through everything above, we have proven true the converse (opposite) of the Isosceles Triangle Theorem.

## Handy Calculations for Isosceles Triangles

In addition to understanding the Isosceles Triangle Theorem, you should also be familiar with a few basic equations for isosceles triangles.

### Height

You can use the Pythagorean theorem to find the height of any triangle. First, label the two equal sides as a, and the base as b. The height (h) equals the square root of b2 – 1/4 a2. You can also divide the square root of 4a2 – b2in half and get the same result.

### Perimeter

To measure the length of the outside of a triangle, add the length of each side together. But, since isosceles triangles have two equal sides, you can make the process easier with this formula: p = 2a + b.

### Area

Area is defined as the total of unit squares you can fit inside any given shape. For an isosceles triangle, divide the total of the base (b) x height (h)by 2.

If you don’t know the height, use the formula listed above to calculate it.

### Altitude

You can find the altitude of the isosceles triangle given the base (B) and the leg (L) by taking the square root of L2 – (B/2)2.

### Base

To find the base of an isosceles triangle when you know the altitude (A) and leg (L), it is 2 x the square root of L2 – A2.

### Leg

When given the base (B) and altitude (A), the leg is the square root of A2 + (B/2)2.

### Interior Angle

When you know one interior angle of an isosceles triangle, it’s possible to find the other two. Let’s say that the angle at the apex is 40 degrees. Because angles must add up to 180 degrees, the two base angles need to add up to 140. Given they must be congruent angles, each of them must be 70 degrees.

## Conclusion

Image via: Flickr

In this article, we have covered the history of isosceles triangles, the different types of triangles, useful formulas, and various applications of isosceles triangles.

We also discussed the Isosceles Triangle Theorem to help you mathematically prove congruent isosceles triangles. For a little something extra, we also covered the converse of the Isosceles Triangle Theorem. You should be well prepared when it comes time to test your knowledge of isosceles triangles.

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## Problem 14: Power Rule for Integration

Looking to figure out how to sole the power rule for integration for your calculus homework? We have the answers at calculus-help.com.

2019-2020

Find the antiderivatives of each with respect to x, without using u-substitution (i.e., without changing variables):

(a)

(b) , assuming a, b, c, d, and e are constants

## Integral Power Rule Solution:

(a) For the integral power rule solution, start by expanding the numerator, squaring (x – 1) and mutliplying by (2x + 1). Rewrite the denominator as a term with a negative exponent, and then distribute it through the quantity.

Now you can antidifferentiate by adding 1 to each power individually and multiplying the coefficient by the reciprocal of the resulting exponent. In other words, for the first term, add 1 to 3/2 to get 5/2. Then, multiply 2 by the reciprocal of 5/2, which is 2/5.

Don’t forget that a general antiderivative (i.e., an indefinite integral) must always contain “+ C.”

(b) Follow the same procedure, beginning by writing the radical term with a fractional exponent. Then,  distribute and find the antiderivtaive. The variables make it marginally more difficult, especially when you add 1 to the weird variable powers.

## Integrals of radical functions

To apply the power rule for integration to this type of function, you have to remember an important rule from algebra.

If we can write the function using exponents then we most likely can apply the power rule.

Let’s solve this problem:

∫ √x+4 dx

Before even using any calculus, you can rewrite the function using the above rule with exponents. So, you have exponents and can apply the power  rule.

It becomes:

∫√x + 4 dx = ∫x^1/2 +4 dx

Then, you can apply the power rule.

It then becomes:

x^1/2 +4 dx= (x^(1/2+1)) / ( 1/2 + 1) + 4x + C

When simplified, you get the final answer.

That is:

(x^3/2 / 3/2) + 4x + C

Which equals:

(2/3 * x^2/3) + 4x + C

The final answer can usually be written with exponents, like we did here. Or, it can be written with roots. However, your teacher or professor may have a preference, so make sure you ask!

## Finding the Power Rule for Integration

As with anything else related to calculus, perfecting the power rule for integration will take a lot of practice if it doesn’t come to naturally.

Make sure you look up practice problems online! You can browse our other calculus problems here. Best of luck!

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## Problem 6: Table of Derivatives

Need help figuring out the table of derivatives in your calculus homework? We have the answers and help you need at calculus-help.com.

## 2019-2020 Derivative Table

Two functions, f(x) and g(x), are continuous and differentiable for all real numbers. Some values of the functions and their derivatives are given in the following table.

Based on that glorious table, calculate the following:

(a)

(b)

(c)

(d)

### Solution:

(a) Take the derivative of each function separately (the derivative of a sum is equal to sum of its derivatives) and plug in 4 to each to get your answer. Reference the chart for the values of f ‘(4) andg(4).

(b) This time you have to use the Product Rule, because f(x) and g(x) are multiplied. Once again, after you apply the derivative rule, just nab the needed function and derivative values from the chart.

(c) This time it’s the Quotient Rule that has to be applied.

(d) How about a big, warm welcome for the Chain Rule! Remember, you apply the Chain Rule when one function is composed with (inside of) another. To differentiate, take the derivative of the outer functionf(x) while leaving g(x) alone inside f(x). Then multiply by the derivative of g(x).

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## Problem 15: A Grizzly Motion Problem

2011-2012 The motion of a grizzly bear stalking its prey, walking left and right of a fixed point in feet per second, can be modeled by the motion of a particle moving left and right along the x-axis, according to the following acceleration equation: Assume that the origin corresponds to the fixed point, and that … Continue reading "Problem 15: A Grizzly Motion Problem"

2011-2012

The motion of a grizzly bear stalking its prey, walking left and right of a fixed point in feet per second, can be modeled by the motion of a particle moving left and right along the x-axis, according to the following acceleration equation:

Assume that the origin corresponds to the fixed point, and that a positive value for position means that the bear is located to the right of the fixed point as we watch said bear from a safe location.

If the bear’s velocity is 1 ft/sec when t = 0, answer the following questions:

(a) Identify the velocity equation that represents the bear’s motion.

(b) Determine how fast the bear was traveling at t = 7 seconds.

(c) In what direction is the bear traveling at t = 5 seconds?

(d) How far does the bear walk during the first 10 seconds?

Note: You can (and should) use a graphing calculator for part (d).

### Solution:

(a) You are given the acceleration equation. Recall that velocity is the antiderivative of acceleration, so integrate a(t) and use the fact that v(0) = 1  to identify the velocity equation.

(b) Evaluate v(7).

Speed is the absolute value of velocity, so the bear is traveling at a speed of 1.579 ft/sec when t = 7.

(c) Evaluate v(5).

Because v(5) is negative, the bear is traveling left at t = 5.

(d) You must split the interval [0,10] into segments based on the t-intercepts of v(t). Those values of t are the times at which the bear changes direction; you must measure how far the bear traveled forward and backward separately.

The velocity equation has only one t-intercept on [0,10]: t = 1.17012095. To calculate the total distance traveled, compute the area between v(t) and the t-axis on the intervals [0,1.17012095] and [1.17012095,10] independently. When the bear is traveling left, this integral is a negative value, but you are asked to find the total distance traveled, not the final position of the bear. Therefore, both integrals must be positive values, so take the absolute value of the second definite integral.

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## Problem 13: Polar Derivatives

2011-2012 Find all angles on the interval  at which the tangent line to the graph of the polar equation  is horizontal. Solution: Express the polar equations parametrically (in terms ofx and y) and calculate the slope of the polar equation. The tangent lines to the polar graph are horizontal when the numerator of this derivative is … Continue reading "Problem 13: Polar Derivatives"

## 2011-2012

Find all angles on the interval  at which the tangent line to the graph of the polar equation  is horizontal.

### Solution:

Express the polar equations parametrically (in terms ofx and y) and calculate the slope of the polar equation.

The tangent lines to the polar graph are horizontal when the numerator of this derivative is equal to 0. In other words, at .

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## Problem 12: Super Related Rates

2011-2012 Have you ever loved something so deeply, so meaningfully, so completely, so profoundly that it would really irk you if you dropped that thing into a bubbling vat of acid? I have, and so that you may learn from my tragedy, I will share a horrific tale from my past. Once, on a whim, … Continue reading "Problem 12: Super Related Rates"

2011-2012

Have you ever loved something so deeply, so meaningfully, so completely, so profoundly that it would really irk you if you dropped that thing into a bubbling vat of acid? I have, and so that you may learn from my tragedy, I will share a horrific tale from my past.

Once, on a whim, I spent an entire summer trying to carve a perfect cube from a piece of driftwood on the beach. Don’t ask why; this is what all math teachers do during summer break, and if teachers tell you otherwise, they are lying. Look at their hands carefully—they are probably whittling as they lie to you!

After months of hard work and risking my fish-belly white skin to the sun, I finally found the perfect piece of driftwood. In just a few days, up to my knees in whittled-away wood chips, I beheld the most beautiful hand-carved wooden cube, each edge precisely one inch long. To say it was a thing of beauty is an understatement. In fact, we became fast friends, and spent much time laughing together about the absurdity of life. (I was later to learn that I had a severe case of sun poisoning and may have hallucinated some of this.)

One day I took my little wooden pal to my favorite acid factory. These were the days of the go-go 80s, when you could just walk around giant open vats of hydrochloric acid and take photos, as long as you either wore a hard hat or styled your hair to look like a hard hat. Intent to give my wooden cube the best possible vantage point to see the massive, bubbling vats of acid, I held him precariously over the safety railings.

What happened next? I think you already know. The cube tumbled from my red, blistered fingers into the acid, and the cube vanished from sight. Moments later, it skyrocketed out of the acid, its edges expanding at an absurd 2 ft/sec. I watched with equal parts pride and horror as the cube also gained super crime-fighting powers (including, but not limited to, the power to fly, fill out tax forms correctly, and chew through solid rock). It even retained its perfect cubic shape as it grew to a monstrous size.

I lost consciousness almost instantly, but two moments are seared into my memory, when its still-growing form had an edge length of 12 ft and (later) when the edge length reached an incredible 210 ft.

(a) At what rate did the surface area of the cube change when its edges were 12 feet long? How about 210 feet long?

(b) At what rate did the volume of the cube change when its edges were 12 feet long? Once again, how about later, when its sides measured precisely 210 feet long?

(c) What was the name of the Super Villain whose life ambition was to fight against my beautiful, yet terrifying, cube?

### Solution:

(a) First, you need the equation for surface area of a cube, and you’ll take the derivative of it with respect to time. The surface area, s, of a cube is equal to 6e2, where e is the length of an edge. Basically, it’s a collection of six squares, each with an area of e2.

According to the problem, de/dt = 2 ft/sec, and you need to calculate ds/dt. Take the derivative of the surface area formula with respect to time.

Notice that I plugged in 2 for de/dt. Now you can calculate the two answers required in part (a) by plugging in each, separately, for e:

• When e = 12, ds/st = 24(12) = 288 ft2/sec.
• When e = 210, ds/dt = 5,040 ft2/sec.

Don’t forget units. It’s unnecessary to carry them all the way through, but at least insert them at the end. You use ft2/sec because area is expressed in square units (square feet, in this case, instead of feet).

(b) Apply a technique similar to part (a), using the formula for the volume V of a cube: V = e3.

Substitute e = 12 and e = 210 into this formula to get answers of 864 ft3/sec and 264,600 ft3/sec, respectively.

(c) The Super Villian’s name was Butter Fingers, but we will also accept Cameron, the Devourer of Worldsas a correct answer.

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## Problem 11: Chain Rule

2011-2012 Chain Rule Problems Calculate the derivative of tan2(2x –1) with respect tox using the chain rule, and then verify your answer using a second differentiation technique. Solution for Chain Rule Practice Problems: Note that tan2(2x –1) = [tan (2x – 1)]2. To find the solution for chain rule problems, complete these steps: Apply the power rule, … Continue reading "Problem 11: Chain Rule"

## 2011-2012 Chain Rule Problems

Calculate the derivative of tan2(2x –1) with respect tox using the chain rule, and then verify your answer using a second differentiation technique.

## Solution for Chain Rule Practice Problems:

Note that tan2(2x –1) = [tan (2x – 1)]2. To find the solution for chain rule problems, complete these steps:

1. Apply the power rule, changing the exponent of 2 into the coefficient of tan (2x – 1), and then subtracting 1 from the square.
2. Multiply by the expression tan (2x – 1), which was originally raised to the second power.
3. Take the derivative of tan (2x – 1) with respect to x.
4. Multiply by the derivative of 2x – 1, the expression that is plugged into tangent.

## Derivative of tan2x

These four steps are implemented in the solution below for derivative of tan2x.

To verify the derivative, apply the product rule, noting that  tan2(2x –1) = tan (2x –1) · tan (2x – 1). Evev the product rule will require the chain rule, when you differentiate each factor (2x – 1), as demonstrated below.

Both techniques for these chain rule practice problems result in the same derivative.

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## Problem 10: Diving into Rates of Change

2011-2012 For the second year in a row, one toy looks to dominate the market once again: My First Cliff Diving Kit. It is all the rage, because it comes with everything you need to be an effective cliff diver: swim trunks, neck brace, legal documents for naming next of kin, and very detailed one-paragraph … Continue reading "Problem 10: Diving into Rates of Change"

2011-2012

For the second year in a row, one toy looks to dominate the market once again: My First Cliff Diving Kit. It is all the rage, because it comes with everything you need to be an effective cliff diver: swim trunks, neck brace, legal documents for naming next of kin, and very detailed one-paragraph instruction sheet about how to cliff dive safely. Below, you’ll find a chart representing the sales figures for the each month it has been available. Use that chart to answer the questions that follow.

(a) Assuming that you can draw a continuous graph connecting all the data points, write the equation of line L, the tangent line to the sales graph when x = May 2011.

(b) What is the average rate of sales between November 2010 and September 2011?

(c) Draw a rough sketch of the graph representing the rate of sales between Oct 2010 and Oct 2011.

### Solution:

(a) To write the equation of a line, you need two things: a point and a slope. You’ll need to assign some x-values, because the independent variables are months and not numbers. Assign October 2010 an x-value of 0 and October 2011 an x-value of 12. This produces the graph below.

Therefore, the point representing May 2011 is the coordinate (7,1559). To estimate the tangent there, calculate the slope between that point and the point immediately preceding it, (6,1509).

Therefore, a good approximation for the tangent line would be y – 1559 = 50(x – 7), according to point-slope form. You could also approximate the derivative by using the point immediately after May 2011, and you’ll get a slope of 128 rather than 50. That seems like a huge difference, but since you don’t have any intermediate points, who’s to know which is the better approximation?

(b) Calculate the average rate of change using the slope of the secant line connecting the points: (1,634) and (11,2445).

Therefore, the kits are selling better overall on the entire time period when compared with how they were selling in May 2011, no matter which derivative approximation from part (a) you use.

(c) Use the same process as part (a) to approximate each derivative, and you’ll get something that looks roughly like this the graph below.

Notice that the graph dips below the x-axis on roughly the interval (2,4) because the data seems to be decreasing there. the faster the data points increase, the greater the height of the blue derivative graph.

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## Problem 9: The Graph of a Derivative

2011-2012 Derivative Graph In the below graph, two functions are pictured, f(x) and its derivative, but I can’t seem to tell which is which. According to those graphs, which is greater: Solution: You might have noticed that the red function has an even degree whereas the blue has an odd degree. Why? An even-degreed function’s ends will … Continue reading "Problem 9: The Graph of a Derivative"

## 2011-2012 Derivative Graph

In the below graph, two functions are pictured, f(x) and its derivative, but I can’t seem to tell which is which.

According to those graphs, which is greater:

### Solution:

You might have noticed that the red function has an even degree whereas the blue has an odd degree. Why? An even-degreed function’s ends will either both point up or both point down (here they both point up at the edges of the graph), whereas a function with an odd degree has ends that go in opposite directions (like the blue graph which goes down to the left but up to the right). Even though this is interesting, it is not enough to answer the question.

You may also say that the red graph is definitely a lesser degree because it has a fewer number of x-intercepts (4) than the blue graph (5), but this, too, is not sufficient information to answer the question correctly. It is true that the blue graph is f(x) and the red is f ‘(x), but for different reasons.

Here are two acceptable justifications for identifying the blue graph as f(x):

• Whenever the blue graph has a relative max or min (hilltop or valley bottom, respectively), the red graph has an x-intercept; remember that the derivative of a function is either 0 or undefined when the original function has a relative extrema point
• Whenever the blue graph is increasing, the red graph is positive (i.e., above the x-axis), and when the blue graph is decreasing, the red is negative

Now that we’ve got that straight, let’s get down to answering the question. You have no idea what the exact function values are, but it turns out that you don’t really need to know them! Since the red graph is decreasing at x = –1/4, then its derivative must be negative there (for the same reason that the second bullet above is true).

What about f (–1)? Because the red graph represents f ‘(x), it’s easy to see that the red graph is just above the x-axis when x = –1. Therefore, f ‘(–1) > 0.

Back to the original question, which asks you to compare two values. You now know that one of those values is positive and the other is negative, so the positive value must be greater.

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2011-2012 A man wants to build a rectangular enclosure for his herd. He only has $900 to spend on the fence and wants the largest size for his money. He plans to build the pen along the river on his property, so he does not have to put a fence on that side. The side … Continue reading "Problem 8: Optimizing a Dirt Farm" ## 2011-2012 A man wants to build a rectangular enclosure for his herd. He only has$900 to spend on the fence and wants the largest size for his money. He plans to build the pen along the river on his property, so he does not have to put a fence on that side. The side of the fence parallel to the fence will cost $5 per foot to build, whereas the sides perpendicular to the river will cost$3 per foot. What dimensions should he choose?

### Solution:

Start by drawing a picture of the situation.

According to the problem, the farmer is attempting to maximize the size of the rectangular plot. Therefore, the primary equation will be area: A = l ∙ w. However, to take the derivative (maximizing the function), you need to eliminate a variable, either l or w; for that, you’ll need a secondary equation.

You know he wants to spend $900. Furthermore, each foot of the 2 fences he’ll use for the width of the yard will cost$3. Each foot of the fence parallel to the river will cost \$5.

Solve the cost equation for one of its variables; I’ll solve for l.

Now plug this in for l in the primary area equation.

Now that you’ve got one variable, maximize by finding the derivative and setting it equal to 0.

To find the corresponding length of the optimized field, plug 75 in for w into the modified secondary equation.

So the optimized field has a fence parallel to the river measuring 90 feet and two other fences connecting it to the river that measure 75 feet each.

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