Optimizing Silver Plating Costs for a Square Box and Athletics Track

In summary, in the first problem, a closed box needs to be constructed with a volume of 36 cm^3 and silver plated on the outside. The cost of plating the top and base is 40 cents per cm^2 and the cost of plating the sides is 30 cents per cm^2. The cost of plating the box can be minimized by finding the values of length and height that produce the largest surface area, which is found by setting the derivative of the surface area equal to zero. The cost of plating the box is $21.82, while the answer key states it should be $21.60. In the second problem, an athletics track has two straight sections and two semi-circular ends
  • #1
clairez93
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0

Homework Statement



1. A closed box of square base and volume 36 cm^3 is to be constructed and silver plated on the outside. Silver plating for the top and the base costs 40 cnets per cm^2 and silver plating for the sides costs 30 cents per cm^2. Calculat ethe cost of plating the box so that the cost is minimized.

2. An athletics track has two "straights" of length l meters and two semi-circular ends of radius x meters. The permieter of the track is 400 m. What values of l and x produce the largest area inside the track? [see attachment #5 for diagram]

Homework Equations





The Attempt at a Solution



1. [tex]V = l^{2}h = 36[/tex]

[tex]SA = l^{2} + 2lh[/tex]

[tex]h = 36/l^{2} [/tex]

[tex]SA = l^{2} + 72l^{-1}[/tex]

[tex]SA' = 2l - 72l^{-2} = 2l - 72/l^{2} = 0[/tex]

[tex]2l^{3} - 72 = 0 [/tex]

[tex]2l^{3} = 72[/tex]

[tex]l = 3.302, w = 3.302, h = 3.302[/tex]

[tex]C = (2lw)(.40) + 2(lh)(.30) + 2(hw)(.30) = $21.82[/tex]

Edit: Latex makes my final cost not clear, it says $21.82.

Answer Key: $21.60


2. [tex]2\pix + 4x + 2l = 400[/tex]

[tex]\pix^{2} + 2lx = A[/tex]

[tex]2l = 400 - 2\pix - 4x[/tex]

[tex]l = 200 - \pix - 2x[/tex]

[tex]\pix^{2}+(400 - 2\pix - 4x)x = A[/tex]

[tex]-\pix^{2} - 4x^{2} + 400x = A[/tex]

[tex]A' = -2\pix - 8x + 400 = 0[/tex]

[tex](-2\pi - 8)x = -400[/tex]

[tex] x = 28.005, l = 56.010[/tex]

Answer Key: x = 63.662, l = 0


I'm not sure what I did wrong.
 

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  • #2
The second one. If the Key Answer is the correct answer. Then I also I solved and achieved that answer.

Try,

Parameter = [tex]400 = 2 \pi r + 2l[/tex]
Area = [tex]\pi r^{2} + 2rl[/tex]

Note: r = x

and work from there.

My answer was [tex]r = \frac{200}{\pi}[/tex] and l = 0.

I didn't have a calculator. But R seems correct.
 
Last edited:
  • #3
Question 1 is probably just a rounding error.
 
  • #4
problem 1

I'm assuming that [itex]SA[/itex] is supposed to represent the total surface area of the box? If so, then shouldn't it be : [tex]SA = 2l^{2} + 4lh[/tex] since the box has a top and bottom and 4 sides?

More importantly, why are you setting the derivative of the surface area equal to zero; wouldn't that minimize/maximize the surface area, not the cost?

You should be minimizing your expression for the cost instead.

problem 2

why is there a [itex]4x[/itex] in your expression for the perimeter ?

general LaTeX pointers

to get your [itex]\pi[/itex]s to show up properly, ou need to put a space between \pi and your next letter: \pi x^2 shows up as [itex]\pi x^2[/itex] but \pix^2 shows up as [itex]\pix^2[/itex]

to display the [itex]\$[/itex] sign use \$ instead of $
 
  • #5
gabbagabbahey said:
problem 1

I'm assuming that [itex]SA[/itex] is supposed to represent the total surface area of the box? If so, then shouldn't it be : [tex]SA = 2l^{2} + 4lh[/tex] since the box has a top and bottom and 4 sides?

More importantly, why are you setting the derivative of the surface area equal to zero; wouldn't that minimize/maximize the surface area, not the cost?

You should be minimizing your expression for the cost instead.

problem 2

why is there a [itex]4x[/itex] in your expression for the perimeter ?

general LaTeX pointers

to get your [itex]\pi[/itex]s to show up properly, ou need to put a space between \pi and your next letter: \pi x^2 shows up as [itex]\pi x^2[/itex] but \pix^2 shows up as [itex]\pix^2[/itex]

to display the [itex]\$[/itex] sign use \$ instead of $

Sorry. But yeah \times looks like the cross product.
 

FAQ: Optimizing Silver Plating Costs for a Square Box and Athletics Track

What is an optimization math problem?

An optimization math problem is a type of mathematical problem that involves finding the maximum or minimum value of a given function, subject to certain constraints. It is commonly used in fields such as engineering, economics, and computer science to help make decisions and improve efficiency.

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To solve an optimization math problem, you first need to define the objective function and any constraints. Then, you can use techniques such as calculus, linear programming, or dynamic programming to find the optimal solution.

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In linear optimization problems, the objective function and constraints are all linear equations. This means that the variables are raised to the first power and there are no products or divisions. Nonlinear optimization problems, on the other hand, involve nonlinear equations and can be more complex to solve.

What are some common challenges in solving optimization math problems?

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