Optimizing Silver Plating Costs for a Square Box and Athletics Track

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Homework Help Overview

The discussion revolves around two optimization problems: one involving the cost of silver plating a closed box with a square base and a specified volume, and the other concerning the dimensions of an athletics track that maximizes the area within a fixed perimeter.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between surface area and cost in the box problem, questioning the setup of the surface area equation and the method of optimization used. There is also discussion about potential rounding errors in the calculations.
  • In the athletics track problem, participants analyze the perimeter equation and area formulation, with some suggesting alternative expressions and questioning the assumptions made regarding the dimensions.

Discussion Status

Some participants have provided insights into the equations used and have raised questions about the correctness of the original poster's approach. There is an ongoing exploration of different interpretations and methods without a clear consensus on the solutions.

Contextual Notes

Participants note potential issues with the original equations and assumptions, including the need for clarification on the surface area calculation for the box and the perimeter expression for the track. There is also mention of formatting issues in LaTeX that could affect clarity.

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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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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:
Question 1 is probably just a rounding error.
 
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 $
 
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.
 

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