Partial derivative and max/min?

Question

A rectangular box must have a volume of 2 cubic meters. The material for the base and top costs $ 5 per square meter. The material for the vertical sides costs $ 160 per square meter.

using 10lw+640(1/w+1/l), find the dimensions of the box that cost the least..

height, width, and length.

Answer 1

First, we need to find the partial derivatives of cost with respect to length and width.

C = 10LW + 640(1/W + 1/L)
dC/dL = 10W - 640 * (1/L^2)
dC/dW = 10L - 640 * (1/W^2)

The minimum of cost will occur where the above partial derivatives are equal to 0.

10W - 640 * (1/L^2) = 0
10W = 640 * (1/L^2)
W(L^2) = 64

10L - 640 * (1/W^2) = 0
10L = 640 * (1/W^2)
L(W^2) = 64

So we have 2 equations which both need to be true at minimum cost

W(L^2) = 64 and L(W^2) = 64
L = 64/(WL) and W = 64/(WL)

So W and L must be the same

W(W^2) = W^3 = 64
W = L = 4 meters
H = 2 cubic meters / (4 * 4 square meters) = 0.125 meters

Note that the given equation is correct, Gary needs to substitute
hw = 2/l and hl = 2/w to get the final result...

Answer 2

Let's say we have l * w * h = 2
The cost of the bottom and top is 10 * l * w, as given.

The cost of the sides is

2 * 160 * w * h + 2 * 160 * l * h = 320h(w + l)

Since h = 2 / (w * l), we get for the cost of the sides

320 * 2 * (w + l) / (w * l) = 640(1/w + 1/l)

Then proceed as HeartSensei

Answer 3

this is one way of observing it: f(x,y) would be best whilst its exponent is best, and smallest whilst its exponent is smallest. of course, the exponent isn't destructive because of the fact you're including 2 squares. yet the two x and y ought to be 0 so which you have got here upon unquestionably the minimum of e^0=one million at x=y=0 Now |y| ? 2 (because of the fact x² + y² ? 4), so the main that x² + 2y² ought to be is (x²+y²) + y² ? 4 + 4, it particularly is accomplished at x=0, y=±2. subsequently, unquestionably the optimal of f(x,y) is e^8 which occurs at (0,±2)