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October 26, 2014

October 26, 2014

Posted by **Jill** on Sunday, January 16, 2011 at 2:03pm.

Romeo is commissioning the construction of a four-inch-tall box made of precious metals to give to his bride, Juliet. The jewelry box will have rectangular sides, a bottom and a top.

The longer sides and the top will be made of gold, at $300/in^2.

The shorter sides will be made of platinum, at $550/in^2.

The bottom will be made of plywood, at $0.02/in^2.

What dimensions will provide Romeo with the lowest cost if he is adamant that the box have a volume of 50 in^3?

I am aware that this involves minimizing the cost, and thus taking the derivative of a cost function for the situation. Can anyone tell me what the cost function for this would be?

Thanks!

- Lagrange multipliers -
**MathMate**, Sunday, January 16, 2011 at 3:09pmLet x=long side, y=short side, z=height

(note: since the cost of material for the short side is more expensive, the minimization will automatically result in y being the shorter side without introducing an additional constraint of x-y>0, but do check when you get your answers)

The cost function would be:

C(x,y,z)

=300xy+2*300xz+2*550yz+0.02xy

=300.02xy+600xz+1100yz

subject to xyz=50.

I get $23,731.91 as the cost.

For your checking purposes, I get 12.64 inches as the sum of x,y and z.

Other Lagrange multiplier examples are available at this site at:

http://www.jiskha.com/display.cgi?id=1289839685

http://www.jiskha.com/display.cgi?id=1291776389

- Calculus -
**Reiny**, Sunday, January 16, 2011 at 3:57pmI think MathMate did not catch the fact that the box has to be 4 inches high, thus eliminating the variable z.

Cost = 300(8x+xy + 550(8y) + .02xy

= 2400x + 300xy + 4400y + .02xy

but V = 4xy

4xy = 50

y = 50/(4x)

cost = 2400x + 300x(50/4x) + 4400(50/(4x) + .02x(50/(4x))

= 2400x + 3750 + 55000/x + .25

d(cost)/dx = 2400 - 55000/x^2 = 0 for min of cost

2400 = 55000/x^2

x^2 = 55000/2400 = 22.916666

x = 4.79

y = 50/(4(4.79)) = 2.61

the box should be 4.79 in long, 2.61 in wide, and 4 inches high

- Oops! -
**MathMate**, Sunday, January 16, 2011 at 5:20pmThanks Reiny!

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