# Optimization problem (Max area of a combined semi circle and a square)

• Mathman2013
In summary: The half-circle area expression is 2*l*w^2. In summary, the figure is made from a semi-circle and square. With the following dimensions, width = w, and length = l. Find the maximum area when the combined perimiter is 8 meter.
Mathman2013
Homework Statement
Optimization problem (Max area of a combined semi circle and a square)
Relevant Equations
f (w) = (4 - 9/7*w)*w + 11/28*w^2
A figure is made from a semi circle and square. With the following dimensions, width = w, and length = l.

Find the maximum area when the combined perimiter is 8 meter.

I first try to construct the a function for the perimeter.

2*l + w + 22/7*w/2 = 8 - > l = 4 - (9*w)/7

Next I insert this into a function which is suppose to define the area.

f(w) = l*w + 1/2*22/7*(w/2)^2 and substitute l with l = 4 - (9*w)/7

which results in f(w) = (4 - 9/7*w)*w + 11/28*w^2

I find the derivative of this function f'(w) = -(25*w)/14 + 4 and solve f'(w) = 0 and find that w is suppose to be 2.24 m

However the solution manual says w = 1.86 m.

So what am I doing wrong?

You forget to make a sketch. What is l and what is w for a square ? and for a circle ?

BvU said:
You forget to make a sketch. What is l and what is w for a square ? and for a circle ?

The drawing in the textbook look like this. Any thoughts on what I am doing wrong? In regards to my calculations?

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• drawing.png
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I can't find what you doing wrong, wolfram also says that w=56/25. My only doubt is regarding the scheme, is that an exact image from the problem at the book?

Thank your answer, the image in the textbook in the included file. The text is in danish, but simply says: A flowerbed is comprised of a semi-circle and square. The perimeter of the entire must be 8 meters in total.

How long does the sides of the square part have to be? To get the maximum area of the square part of flowerbed?

And I constructed a function for the perimeter in initial post post. Calling the length l and width w. Next isolated with respect to l, and next found the area function. f(w) which calculated the derivative of. and solved the equation f'(w) = 0 and found w and used this result to find L.

Can it be an error in the problem? Or an error in the solution manual?

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• flower.jpg
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unfortunately I don't know Danish to read for myself, but does it say "the maximum area of the square part" of flowerbed?. IF so I think it asks to maximize the area of the rectangle with sides l and w, that is to maximize ##l\cdot w##, ignoring the area of the circular part.

Thank you again for your answer, no danish is a small language :)

I try again,
But if the function for the perimeter is 2*l+w + pi *w/2 = 8,(if we assume l is length of the square part, and w is the width) so that l = 4 - (9*w)/7

If I insert this into the function f(w) = w*( 4 - (9*w)/7) - > f(w) = 4*w- (9*w^2)/7

and solve via the derivative, then I get. solve(f'(w) = 0) , then w = 1.555555556 and l = 2.

Which should according to my math give me the optimal values for l and w.

Any thoughts on my calculations?

I think you are correct but we still far from the solution of 1.86 given by the textbook...
I don't know what else to think of, sorry.

You confused things by saying "square". If w is not equal to l, that is not a square. Taking w to be the length of the top and bottom of the rectangle and l to be the length of the right and left sides, then the circle has radius w/2. The distance around the semicircle is $\pi w/2$ and the distance around the three sides of the rectangle is 2l+ w. So the total distance around the figure is $\frac{\pi w}{2}+ 2l+ w= 8$. The area of the semicircle is $\frac{\pi w^2}{4}$ and the area of the rectangle is $lw$. The problem is asking you to maximize $\frac{\pi w^2}{4}+ lw$ subject to the condition that $\frac{\pi w}{2}+ 2l+ w= 8$. The simplest way to do that is to solve $\frac{\pi w}{2}+ 2l+ w= 8$ for l so that you can write $\frac{pi w^2}{4}+ lw$ as a function of w only.

Dear HallsoftIvy,

If I try to solve in Maple, then I get that w is 4?

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• Maple.png
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Mathman2013 said:
semi-circle and square.
Should be rectangle. 2.24 is correct, book answer wrong.

Mathman2013 said:
If I try to solve in Maple, then I get that w is 4?
Check the half-circle area expression...

## 1. What is an optimization problem?

An optimization problem is a mathematical problem that involves finding the maximum or minimum value of a function, subject to various constraints. In other words, it is the process of finding the best possible solution to a given problem.

## 2. What is the "Max area of a combined semi circle and square" optimization problem?

The "Max area of a combined semi circle and square" optimization problem involves finding the maximum possible area that can be enclosed by both a semi circle and a square, given a fixed perimeter. This problem can be used in various real-life situations, such as determining the maximum area for a playground or garden within a given fence length.

## 3. How do you approach solving this optimization problem?

To solve the "Max area of a combined semi circle and square" optimization problem, we use the derivative of the area function with respect to the variable we are trying to optimize (in this case, the radius of the semi circle). We then set the derivative equal to 0 and solve for the optimal value. We also need to check the endpoints of the feasible region to ensure that we have indeed found the maximum value.

## 4. What are the constraints in this optimization problem?

The constraints in the "Max area of a combined semi circle and square" optimization problem are the perimeter and the fact that the area must be positive. The perimeter constraint ensures that the shapes fit within a given boundary, while the positivity constraint ensures that we are dealing with physically possible shapes.

## 5. Can this optimization problem be extended to other shapes?

Yes, the concept of finding the maximum area or volume for a given perimeter or surface area can be extended to other shapes, such as rectangles, triangles, and polygons. The same approach of using derivatives and constraints can be applied to these problems as well.

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