Optimization Problems: Fifty feet of wire is to be used to form two figures....

[Frenchi33]

Fifty feet of wire is to be used to form two figures. In each of the following cases, how much wire should be used for each figure so that the total enclosed area is a maximum?

a) Equilateral triangle and square
b) Square and regular pentagon
c) Regular pentagon and regular hexagon
d) Regular hexagon and circle

I'm kind of stumped on these problems. All I know is that I need a Perimeter Formula, an Area Formula, and will need to take the derivative of the Area Formula at some point during the problem. Can someone please help me? The answers are supposed to be in "ft."
Thank you.

 

[JeffM]

Asked and answered

http://mymathforum.com/calculus/339950-optimization-problems.html

 

[ksdhart2]

The basic outline of steps on these problems will be as follows:

1. Label the length of one side of one shape with a variable, and the length of one side of the other shape with a different variable
2. Create an expression for the perimeter of the two shapes involved. This is made easier by the fact that all the shapes you're working with are regular.
3. Add these together to get a formula for the total perimeter
4. Create an expression for the area of the shapes involved. Again, the fact that all the shapes are regular helps out a lot.
5. Add these together to get a formula for the total area
6. Set the perimeter expression equal to 50 ft
7. Use the equation for the perimeter to solve for one variable in terms of the other
8. Plug that value into the area formula to "reduce" it to an expression of one variable
9. Find the derivative of the new area formula with respect to its variable
10. Set the derivative equal to zero to find any maxima
11. Return to the perimeter equation to solve for the value of the other variable

I'll work an example with a regular octagon and a regular decagon.

1. Let the length of one side of the octagon be o and the length of one side of the decagon be d
2. The perimeter of a regular octagon is 8o, and the perimeter of a regular decagon is 10d.
3. P = 8o + 10d
4. The area of a regular octagon is \(\displaystyle 2(1+\sqrt{2}) \cdot o^2\), and the area of a regular decagon is \(\displaystyle \dfrac{5}{2} \sqrt{5+2\sqrt{5}} \cdot d^2\)
5. \(\displaystyle A = 2(1+\sqrt{2}) \cdot o^2 + \dfrac{5}{2} \sqrt{5+2\sqrt{5}} \cdot d^2\)
6. 8o + 10d = 50
7. 10d = 50 - 8o ; d = 5 - 4o/5
8. \(\displaystyle A = 2(1+\sqrt{2}) \cdot o^2 + \dfrac{5}{2} \sqrt{5+2\sqrt{5}}\cdot \left(5-\dfrac{4o}{5}\right)^2\)
9. \(\displaystyle \dfrac{dA}{do} = 4(1+\sqrt{2}) \cdot o - 4 \left(\sqrt{5+2\sqrt{5}}\right)\cdot \left(5-\dfrac{4o}{5}\right)\)
10. \(\displaystyle 4(1+\sqrt{2}) \cdot o - 4 \left(\sqrt{5+2\sqrt{5}}\right)\cdot \left(5-\dfrac{4o}{5}\right) = 0 \implies o \approx 3.1557\)
11. \(\displaystyle 50 = 8o + 10d \implies 50 = 8(3.1557) + 10d \implies d \approx 2.47544\)

This result says that, to maximize the enclosed area, you'd want to use about 25.25 feet on the octagon and the remaining 24.75 feet on the decagon. To solve the problems, you may need to know the formulas for the area of a regular pentagon and a regular hexagon. You can either look these up on Google (they may also be in your textbook somewhere) or derive them from the general formulahttp://www.mathwords.com/a/area_regular_polygon.htm for the area of a regular polygon:

\(\displaystyle A = \dfrac{1}{4} ns^2 \cdot cot \left( \dfrac{\pi}{n} \right)\)

where n is the number of sides and s is the length of one side

 

[nbg273]

The basic outline of steps on these problems will be as follows:

1. Label the length of one side of one shape with a variable, and the length of one side of the other shape with a different variable
2. Create an expression for the perimeter of the two shapes involved. This is made easier by the fact that all the shapes you're working with are regular.
3. Add these together to get a formula for the total perimeter
4. Create an expression for the area of the shapes involved. Again, the fact that all the shapes are regular helps out a lot.
5. Add these together to get a formula for the total area
6. Set the perimeter expression equal to 50 ft
7. Use the equation for the perimeter to solve for one variable in terms of the other
8. Plug that value into the area formula to "reduce" it to an expression of one variable
9. Find the derivative of the new area formula with respect to its variable
10. Set the derivative equal to zero to find any maxima
11. Return to the perimeter equation to solve for the value of the other variable

I'll work an example with a regular octagon and a regular decagon.

1. Let the length of one side of the octagon be o and the length of one side of the decagon be d
2. The perimeter of a regular octagon is 8o, and the perimeter of a regular decagon is 10d.
3. P = 8o + 10d
4. The area of a regular octagon is \(\displaystyle 2(1+\sqrt{2}) \cdot o^2\), and the area of a regular decagon is \(\displaystyle \dfrac{5}{2} \sqrt{5+2\sqrt{5}} \cdot d^2\)
5. \(\displaystyle A = 2(1+\sqrt{2}) \cdot o^2 + \dfrac{5}{2} \sqrt{5+2\sqrt{5}} \cdot d^2\)
6. 8o + 10d = 50
7. 10d = 50 - 8o ; d = 5 - 4o/5
8. \(\displaystyle A = 2(1+\sqrt{2}) \cdot o^2 + \dfrac{5}{2} \sqrt{5+2\sqrt{5}}\cdot \left(5-\dfrac{4o}{5}\right)^2\)
9. \(\displaystyle \dfrac{dA}{do} = 4(1+\sqrt{2}) \cdot o - 4 \left(\sqrt{5+2\sqrt{5}}\right)\cdot \left(5-\dfrac{4o}{5}\right)\)
10. \(\displaystyle 4(1+\sqrt{2}) \cdot o - 4 \left(\sqrt{5+2\sqrt{5}}\right)\cdot \left(5-\dfrac{4o}{5}\right) = 0 \implies o \approx 3.1557\)
11. \(\displaystyle 50 = 8o + 10d \implies 50 = 8(3.1557) + 10d \implies d \approx 2.47544\)

This result says that, to maximize the enclosed area, you'd want to use about 25.25 feet on the octagon and the remaining 24.75 feet on the decagon. To solve the problems, you may need to know the formulas for the area of a regular pentagon and a regular hexagon. You can either look these up on Google (they may also be in your textbook somewhere) or derive them from the general formula for the area of a regular polygon:

\(\displaystyle A = \dfrac{1}{4} ns^2 \cdot cot \left( \dfrac{\pi}{n} \right)\)

where n is the number of sides and s is the length of one side

Thank you, but I'm still not getting the right answer. For (a) with the equilateral triangle and square problem, this is what I've done so far...

1) have the formulas - 3s+4x=50, A=(s^(2)sqrt(3))/(4)+x^(2)

2) solved for s and plugged it into the area formula - A=(((50/3)-(4x/3))^(2)sqrt(3))/(4)+x^(2)

3) took the derivative and solved for x - A'=(4(-2x+25)/3(sqrt(3))+2x, x=[5.437056467] which isn't right on my online homework, so I'm obviously not going to solve for s yet.

What am I doing wrong? Thank you.

 

[JeffM]

Thank you, but I'm still not getting the right answer. For (a) with the equilateral triangle and square problem, this is what I've done so far...

1) have the formulas - 3s+4x=50, A=(s^(2)sqrt(3))/(4)+x^(2)

2) solved for s and plugged it into the area formula - A=(((50/3)-(4x/3))^(2)sqrt(3))/(4)+x^(2)

3) took the derivative and solved for x - A'=(4(-2x+25)/3(sqrt(3))+2x, x=[5.437056467] which isn't right on my online homework, so I'm obviously not going to solve for s yet.

What am I doing wrong? Thank you.

I am somewhat confused here. Partly because I have no idea what you know; partly because I am not sure you have given the problem completely and exactly, and partly because you gave no working on your solution so finding an error is well nigh impossible.

Your initial formulas for the sum of the areas and perimeters are fine.

\(\displaystyle a = x^2 + \dfrac{s^2\sqrt{3}}{4} \text { and } 50 = p = 4x + 3s \implies s = \dfrac{50 - 4x}{3} \implies s^2 = \dfrac{2500 - 400x + 16x^2}{9}.\)

And your solution for s in terms of x is good, but I think you are setting yourself up for trouble by failing to simplify the formula for the sum of the areas.

\(\displaystyle a = x^2 + \dfrac{\sqrt{3}}{4} * \dfrac{2500 - 400x + 16x^2}{9} = \dfrac{1}{36} * \{ 2500\sqrt{3} - 400x\sqrt{3} + x^2(36 + 16\sqrt{3}) \} \implies \)

\(\displaystyle a' = \dfrac{1}{36} * \{ 2x(36 + 16\sqrt{3}) - 400\} = \dfrac{1}{18} * \{ x(36 + 16 \sqrt{3}) - 200\}.\)

So I do not get your derivative. But it hardly makes any difference because the formula for the sum of the areas is a quadratic with a positive leading coefficient, which doesn't have a maximum anywhere.

What does the problem say exactly and completely? Are you studying boundary conditions?

 

[nbg273]

I am somewhat confused here. Partly because I have no idea what you know; partly because I am not sure you have given the problem completely and exactly, and partly because you gave no working on your solution so finding an error is well nigh impossible.

Your initial formulas for the sum of the areas and perimeters are fine.

\(\displaystyle a = x^2 + \dfrac{s^2\sqrt{3}}{4} \text { and } 50 = p = 4x + 3s \implies s = \dfrac{50 - 4x}{3} \implies s^2 = \dfrac{2500 - 400x + 16x^2}{9}.\)

And your solution for s in terms of x is good, but I think you are setting yourself up for trouble by failing to simplify the formula for the sum of the areas.

\(\displaystyle a = x^2 + \dfrac{\sqrt{3}}{4} * \dfrac{2500 - 400x + 16x^2}{9} = \dfrac{1}{36} * \{ 2500\sqrt{3} - 400x\sqrt{3} + x^2(36 + 16\sqrt{3}) \} \implies \)

\(\displaystyle a' = \dfrac{1}{36} * \{ 2x(36 + 16\sqrt{3}) - 400\} = \dfrac{1}{18} * \{ x(36 + 16 \sqrt{3}) - 200\}.\)

So I do not get your derivative. But it hardly makes any difference because the formula for the sum of the areas is a quadratic with a positive leading coefficient, which doesn't have a maximum anywhere.

What does the problem say exactly and completely? Are you studying boundary conditions?

So I don't know that much, as I'm trying to learn calculus for myself and am using this online homework thing. I guess I don't know how to maximize which is what these problems are about. I'm really just looking for the answer for the mean while, cause it's been driving me crazy for 2 weeks. It sounds dumb, but I learn from getting answers and then backtracking through the problem. That way I know I'm doing something right if I have the answer already. I know people on the forums hate giving people answers, cause they assume everyone's cheating or something...

Anyways, so...

A'=(1/18)*{x(36+16sqrt(3))-200}

I need to solve this for x, right? And then plug the x value into the original p=4x+3s formula? I have no clue what maximizing means.

 

[JeffM]

So I don't know that much, as I'm trying to learn calculus for myself and am using this online homework thing.

I have neither the time nor the energy nor the capacity to write a text on calculus. I suggest you watch the YouTube videos at Khan Academy on calculus from the beginning to supplement your online homework thing. But it is good to know that you are a beginning student so that we do not assume you know a great deal.

I guess I don't know how to maximize which is what these problems are about.

One of the major purposes of differential calculus is to find maximum and minimum values of a function. Such a value is called an extremum (plural is extrema). Finding maximum values is called maximization. Finding minimum values is called minimization. Finding extrema is called optimization. Clear so far?

A global maximum is the highest value a function can attain anywhere in its domain. A global minimum is the lowest value a function can attain anywhere in its domain. It is important to note that a function may or may not have a global maximum and may or may not have a global minimum. The function \(\displaystyle x^3\) has neither a global minimum nor a global maximum. The function \(\displaystyle x^2\) has a global minimum but not a global maximum. The function \(\displaystyle -\ x^2\) has a global maximum but not a global minimum. The function \(\displaystyle sin(x)\) has has both a global maximum and a global minimum.

Do you have this?

If f(x) is differentiable and has a global maximum at a, that function's derivative is zero at a. If f(x) is differentiable and has a global minimum at b, that function's derivative is zero at b. Warning: a function's derivative may be zero at c without the function having a minimum or a maximum there. So finding that a function's derivative is zero at c does NOT tell you that the function has a global maximum or a global minimum at c. You have to do additional tests. Still with me?

Ok. So the process of finding the global maximum (minimum) for f(x) is to: (1) derive f'(x), the derivative of f(x), (2) determine what values of x make f'(x) = 0, (3) test whether any of those values (if there are any) is a global maximum (minimum).

Before continuing, make sure you understand all this. Ask whatever questions you have about the preceding. Check out Khan Academy. Then we can proceed.