# Optimization Problem

• Apr 8th 2007, 06:12 PM
zachb
Optimization Problem
I need help with two optimization problems. The first one involves the volume of a box and I have no idea what the second problem is asking.

1) If 1200 cm^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box?

2) (a) Show that of all the rectangles with a given area, the one with smallest perimeter is a square.

(b) Show that of all rectangles with a given perimeter, that one with greatest area is a square.
• Apr 8th 2007, 06:53 PM
Jhevon
Quote:

Originally Posted by zachb
I need help with two optimization problems. The first one involves the volume of a box and I have no idea what the second problem is asking.

1) If 1200 cm^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box?

See diagram below:

Let S be the surface area
Let V be the volume of the box

We have S = 1200
=> 1200 = x^2 + 4xy .........see diagram
=> y = 300/x - x/4

Now V = x^2 * y
=> V = x^2 (300/x - x/4)
=> V = 300x - (x^3)/4
=> V' = 300 -(3/4)x^2

For max V, set V' = 0
=> 300 -(3/4)x^2 = 0
=> (3/4)x^2 = 300
=> x^2 = 400
=> x = +/- 20
=> x = 20, since x cannot be negative

but y = 300/x - x/4
=> y = 300/20 - 20/4
=> y = 10

so for max volume, the dimensions are x = 20, y = 10

since V = x^2 * y
max V = (20)^2 * 10
=> maxV = 4000 cm^3
• Apr 8th 2007, 07:03 PM
Jhevon
Quote:

Originally Posted by zachb
2) (a) Show that of all the rectangles with a given area, the one with smallest perimeter is a square.

(b) Show that of all rectangles with a given perimeter, that one with greatest area is a square.

(a) Let the dimensions of the rectangle be length = x and width = y.

then the given area, A = xy
=> y = A/x

now the perimeter P of the rectangle will be given by
P = 2x + 2y
=> P = 2x + 2(A/x) = 2x + 2A*x^-1
=> P' = 2 - 2A*x^-2
for smallest P, set P' = 0
=> 2 - 2A*x^-2 = 0
=> 2A*x^-2 = 2
=> x^-2 = 1/A ............divided both sides by 2A
=> x^2 = A ................took the inverse of both sides
=> x = +/- sqrt(A)

but y = A/x
=> y = +/- A/sqrt(A)
=> y = +/- sqrt(A)

so we see that y = x = +/- sqrt(A), so our desired rectangle is actually a square

b) This is done similar to part (a), try it and post your solution and i'll guide you if you go wrong
• Apr 8th 2007, 08:09 PM
zachb
P = 2x + 2y
=> y = P - 2x/2 or P/2 - x

A = xy
=> A = x(P - 2x/2)
=> A = (px - 2x^2)/2
=> A' = -2x (I'm not sure that A' is correct)

I'm not sure what to do after this point. I set A' = 0, but I don't think that's what I'm supposed to do here. I worked it all the way through but it was not a square.
• Apr 8th 2007, 08:17 PM
Jhevon
Quote:

Originally Posted by zachb
P = 2x + 2y
=> y = P - 2x/2 or P/2 - x

A = xy
=> A = x(P - 2x/2)
=> A = (px - 2x^2)/2
=> A' = -2x (I'm not sure that A' is correct)

I'm not sure what to do after this point. I set A' = 0, but I don't think that's what I'm supposed to do here. I worked it all the way through but it was not a square.

ok, let's see what happened.

we are given P = 2x + 2y
solve for y we obtain, y = (P - 2x)/2 = P/2 - x

Now the area will be given by
A = xy
=> A = x(P/2 - x)
=> A = (P/2)x - x^2

=> A' = P/2 - 2x .........ah, so your A' was not correct, that's the problem. anyway, let's continue

for max A, set A' = 0
=> P/2 - 2x = 0
=> 2x = P/2
=> x = P/4

but y = P/2 - x
=> y = P/2 - P/4
=> y = P/4

so we see x = y = P/4, so our desired rectangle is a square
• Apr 8th 2007, 08:30 PM
zachb
Okay, that make sense and I know why I got A' wrong. Thanks.