Optimization math Problems

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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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Answers and Replies

  • #2
85
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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:
  • #3
85
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Question 1 is probably just a rounding error.
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
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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
85
0
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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