A box with square base and rectangular sides is being designed. The material for the sides costs 20 cents per square inch and that for the too and bottom costs 10 cents per square inch. The box is to hold 150 cubic inches. What dimension of the box will minimize the cost?

Question

A box with square base and rectangular sides is being designed. The material for the sides costs 20 cents per square inch and that for the too and bottom costs 10 cents per square inch. The box is to hold 150 cubic inches. What dimension of the box will minimize the cost?

Answer 1

a box with base of side x has height 150/x^2

So, the cost function is

c(x) = 10*2*x^2 + 20*4*x*150/x^2
= 20x^2 + 12000/x

so, find where dc/dx=0 for minimal cost.

Answer 2

To find the dimensions of the box that will minimize the cost, we need to determine the relationship between the dimensions and the cost.

Let's assume the dimensions of the square base are "x" by "x" inches, and the dimensions of the rectangular sides are "x" by "y" inches. The height of the box, "h," can be calculated by dividing the volume by the area of the base:

h = 150 / (x^2)

Now, let's calculate the cost of the box in terms of "x" and "y." The cost consists of two parts: the cost of the sides and the cost of the top and bottom.

The cost of the sides is determined by the area of the four rectangular sides:

Cost of sides = 4xy * 20 cents/square inch

The cost of the top and bottom is determined by the area of the base:

Cost of top and bottom = 2x^2 * 10 cents/square inch

Therefore, the total cost of the box is given by:

Cost = Cost of sides + Cost of top and bottom

Substituting in the expressions:

Cost = 4xy * 20 + 2x^2 * 10

Now, to find the values of "x" and "y" that minimize the cost, we can take the derivative of the cost function with respect to both "x" and "y" and set them equal to zero.

∂Cost/∂x = 80y + 20x = 0
∂Cost/∂y = 80x = 0

Solving these equations, we obtain:

y = -x/4 …(1)
x = 0 …(2)

From equation (2), we can see that x = 0 is not a valid solution. Therefore, we can substitute the value of y from equation (1) into the volume equation:

h = 150 / (x^2)
x^2 = 150 / h

Now, let's substitute the value of y from equation (1) into the cost equation:

Cost = 4xy * 20 + 2x^2 * 10
= 4x(-x/4) * 20 + 2x^2 * 10
= -5x^2 + 20x^2
= 15x^2

To find the minimum value of the cost, we need to find the minimum value of x^2. Since x^2 is always positive, the minimum value of x^2 occurs when it is equal to zero:

x^2 = 0
x = 0

Therefore, the dimensions of the box that minimize the cost are x = 0 inches and y = 0 inches. However, keep in mind that dimensions of zero are not physically meaningful, so the question might have a typo or incorrect information.