# A company wishes to manufacture a box with a volume of 40 cubic feet that is open on top and is twice as long as it is wide.Find the width of the box that can be produced using the minimum amount...

A company wishes to manufacture a box with a volume of 40 cubic feet that is open on top and is twice as long as it is wide.

Find the width of the box that can be produced using the minimum amount of material.

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Let the width of the box be W. As given in the problem, its length is twice the width or is equal to 2W.

Now the volume of the box is length * width * height

= W * 2W * height = 40

Therefore we can express the height as 40 / 2W^2 = 20/W^2.

Now the area of the material required to make the box is the sum of the sides and the base. ( As the top is open)

So the total area is W*2W + 2*W*[20 / W^2] + 2*2*W*[20 / W^2] = 2W^2 + 40/W + 2*40/W = 2W^2 + 3*40/W

Now we have the area of the material required as an expression with a single variable: 2W^2+ 3*40/W

We need to find the minimum value of this expression. So we differentiate it and equate the derivative to 0.

=> 4W - (3*40)/W^2 =0

=> 4W = (3*40)/W^2

=> W^3 = 30

=> W = 30^(1/3)

**The width to make the box with the least amount of material is 30^(1/3).**

Let x and 2x and h be the width ,linth and height of the box.

Then the volume of the box = length*width*height = x*2x*h =2x^2h. But 2x^2*h = 40 cubic feet,

Therefore h = 40/2x^2 = 20/x^2.

Now the material of the box whose sides are x , 2x and h is proportional to its surface area which is x*2x+2x*h+2(2x)h = 2x^2+6xh = x^2+6*(20/x^2) = x^2+120/x

So we shoild minimum area or the minimum value of x^2+120/x.

Let A(x) = 2x^2+6/x. By calculus, minimum A(x) could be A(c), where c is a solution of A'(x) = 0 and makes A''(c) > 0.

A'(x) = 0 gives: (2x^2+120/x)' = 0. Or

4x -120/x^2 = 0. Multiply by 4.

4x^3 = 120

x = 120/4 = 30

x = 30^(1/3).

Also f"(x) = (4x-120/x^2)' = 4-120* (-1/2x^3) = 4+60/x^3 which is > 0 when x=c = (30)^(1/3)

Therefore the minimum material is when the the width = 30^(1/3).