Posted by Jill on Sunday, January 16, 2011 at 2:03pm.
Here is the problem I am working on:
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?
Lagrange multipliers - MathMate, Sunday, January 16, 2011 at 3:09pm
Let 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:
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:
Calculus - Reiny, Sunday, January 16, 2011 at 3:57pm
I 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 5:20pm
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