# A box is contructed out of two different types of metal. The metal for the top and bottom, which are both square, costs \$5 per square foot and the metal for the sides costs \$2 per square foot. Find the dimensions that minimize cost if the box has a volume of 40 cubic feet.

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## To find the dimensions that minimize cost, we need to find the dimensions that minimize the surface area of the box while maintaining a volume of 40 cubic feet.

Let's assume the length of the top and bottom sides of the box is x, and the height of the box is y.

The volume of the box is given as 40 cubic feet, so we have the equation:

x^2 * y = 40

To minimize the cost, we need to minimize the surface area of the box. The surface area is comprised of the top and bottom sides (which are both square and have area x^2) and the four sides, each with an area xy.

Therefore, the surface area of the box is:

2 * x^2 + 4 * xy

Now, we can express the cost of the box in terms of the dimensions:

Cost = (5 * x^2) + (2 * 4 * xy)
Cost = 5x^2 + 8xy

Now, we can use the volume equation to express y in terms of x:

y = 40 / x^2

Substituting this into the cost equation, we get:

Cost = 5x^2 + 8x(40 / x^2)
Cost = 5x^2 + 320 / x

To find the minimum cost, we need to find the value of x that minimizes the cost. Taking the derivative with respect to x and setting it equal to zero will help us find the critical values:

d(Cost) / dx = 10x - 320 / x^2 = 0

Multiplying through by x^2, we get:

10x^3 - 320 = 0

Dividing through by 10, we get:

x^3 - 32 = 0

Taking the cube root of both sides, we get:

x = ∛32 = 2

Therefore, the value of x that minimizes the cost is 2.

Substituting this value back into the equation for y, we get:

y = 40 / (2^2) = 10

So, the dimensions that minimize cost while maintaining a volume of 40 cubic feet are a length and width of 2 feet, and a height of 10 feet.

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## To find the dimensions that minimize the cost of the box, we need to set up an equation for the cost in terms of the dimensions of the box and then minimize that equation.

Let's consider the dimensions of the box:

Let x be the length of the sides (which is the same for the top and bottom) in feet.
Let y be the height of the box in feet.

The volume of the box is given as 40 cubic feet, so we have the equation:

x^2 * y = 40

Next, we need to set up the equation for the cost of the box:

Cost = (2 * x^2 + 4 * x * y) * 2 + x^2 * 2 * 5

The first term (2 * x^2 + 4 * x * y) * 2 represents the cost of the sides of the box (since there are two sides of each dimension), and the second term x^2 * 2 * 5 represents the cost of the top and bottom of the box.

Simplifying the equation:

Cost = 4x^2 + 8xy + 4x^2 + 10x^2
= 14x^2 + 8xy

Now, we have the cost equation in terms of x and y, and the volume equation in terms of x and y. We can substitute the volume equation into the cost equation to get the cost in terms of a single variable (either x or y) and then minimize it.

Substituting the volume equation into the cost equation:

Cost = 14x^2 + 8xy
= 14x^2 + 8x(40/x^2)
= 14x^2 + 320/x

To minimize the cost, we need to find the critical points, meaning where the derivative of the cost equation is equal to zero.

Taking the derivative of the cost equation with respect to x:

d(Cost) / dx = 28x - 320/x^2

Setting the derivative equal to zero and solving for x:

28x - 320/x^2 = 0
28x = 320/x^2
28x^3 = 320
x^3 = 320 / 28
x^3 = 40/7
x = (40/7)^(1/3)

Now that we have the value of x, we can substitute it back into the volume equation to find the corresponding value of y:

x^2 * y = 40
((40/7)^(1/3))^2 * y = 40
(40/7)^(2/3) * y = 40
y = 40 / (40/7)^(2/3)
y = 40 * (7/40)^(2/3)

So, the dimensions that minimize the cost of the box with a volume of 40 cubic feet are approximately:

x ≈ (40/7)^(1/3)
y ≈ 40 * (7/40)^(2/3)

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## So, the height is 40/x^2

c(x) = 2x^2*5 + 4x(40/x^2)*2
= 10x^2 + 320/x

set the derivative to zero, and you find minimum cost at x = 2∛2