A box of volume 72m^3 with square bottom and no top constructed out of two different materials. The cost of the bottom is $40/m^2 and the cost of the sides is $30/m^2. Find the dimensions of the box that minimize the total cost.

How do I set up the equation(s) in the beginning and what is the derivative?

Let

h=height of box,
s=length=width of bottom

Constraint:
Volume, V = hs² = 72

Cost:
Cost, C = 40s²+4*30sh

Let λ=Lagrange multiplier
Objective function, F
F(s,h,λ)=40s²+4*30sh + λ(hs²-72)

Find partial derivatives with respect to each of the three variables and solve for s and l.

I get 3*4^(1/3) for s, and h = 2s/3.

Check my arithmetic.

Trying this without using Lagrange multipiers.

Constraint:
Volume, V = hs² = 72

Cost:
Cost, C = 40s²+4*30sh

so
h = 72/s^2
C = 40 s^2 + 120 s(72/s^2)
C = 40 s^2 + 8640/s
dC/ds = 0 at min = 80 s - 8640/s^2
so
80 s^3 = 8640 at min
s^3 = 108 = 27 * 4
so s = 3 * 4^(1/3) agrees with Math Mate
s = 4.76

To set up the equation, let's assume the side length of the square bottom is x and the height of the box is y. Since the box has no top, the total volume can be expressed as:

V = x^2 * y

The cost of the bottom can be calculated by multiplying the area of the bottom (x^2) by the cost per square meter, which is $40. The cost of the sides can be calculated by multiplying the area of the sides by the cost per square meter, which is $30. The area of the sides is given by:

A = 4(x * y)

Since there are four sides of equal area, each measuring x * y.

The total cost, C, can be expressed as the sum of the cost of the bottom and the cost of the sides, which is:

C = 40(x^2) + 30(4xy)
C = 40x^2 + 120xy

Now, to find the dimensions that minimize the total cost, we need to find the critical points of C. To do this, we can take partial derivatives of C with respect to x and y.

∂C/∂x = 80x + 120y
∂C/∂y = 120x

Setting both derivatives equal to zero and solving for x and y gives us the critical point(s). We can differentiate ∂C/∂x with respect to x to find the value of x that minimizes C, and similarly differentiate ∂C/∂y to find the value of y.

∂C/∂x = 80x + 120y = 0
∂C/∂y = 120x = 0

Solving the first equation, we have:

80x + 120y = 0
8x + 12y = 0
4x + 6y = 0
2x + 3y = 0

Solving the second equation, we have:

120x = 0
x = 0

Since we cannot have a box with a side length or height of 0, we can disregard this solution.

Therefore, the critical point occurs when 2x + 3y = 0, or when:

2x = -3y
x = -(3/2)y

The dimensions of the box that minimize the total cost are given by x = -(3/2)y.

To set up the equation to minimize the total cost, we first need to determine the dimensions of the box. Let's assume the dimensions of the square base are x by x, and the height of the box is h.

The volume of the box is given as 72 m^3, so we can write our first equation based on the volume:
Volume of the box = x * x * h = 72

Next, we need to find an equation for the cost of the bottom and sides. The cost of the bottom is $40/m^2, and the cost of the sides is $30/m^2. The bottom area is x * x = x^2, and the area of each side is 4 * x * h since there are four sides. Therefore, our cost equation is:
Cost = (40 * x^2) + (30 * 4 * x * h)

To minimize the total cost, we need to express the cost equation in terms of a single variable. We can do this by substituting the value of h from the volume equation into the cost equation.

From the volume equation, we have:
h = 72 / (x * x)

Substituting this into the cost equation, we get:
Cost = (40 * x^2) + (30 * 4 * x * (72 / (x * x)))

Let's simplify this equation:
Cost = 40x^2 + 8640 / x

Now we have our cost equation in terms of a single variable, x.

To minimize the cost, we need to find the derivative of the cost equation with respect to x. The derivative will help us find the critical points where the cost is minimized.

Taking the derivative of the cost equation, we get:
dCost/dx = (80x - 8640 / x^2)

Now, the derivative is (80x - 8640 / x^2).

This derivative represents the rate of change of cost with respect to x.