# A company wishes to manufacture a box with a volume of 44 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 minimumamount of material.

The company wishes to manufacture a box with a volume of 44 cubic feet that is open on top and is twice as long as it is wide.

Let the length of the box that uses minimum amount of material be L. As the width of the box is half...

The company wishes to manufacture a box with a volume of 44 cubic feet that is open on top and is twice as long as it is wide.

Let the length of the box that uses minimum amount of material be L. As the width of the box is half the length, it is equal to L/2. If the height of the box is H, `H*L*(L/2) = 44`

=> `H = 88/L^2`

The surface area of the box is equal to `L*(L/2) + 2*L*88/L^2 + 2*(L/2)*88/L^2`

= `L^2/2 + 176/L + 88/L `

= `L^2/2 + 264/L`

To minimize `A = L^2/2 + 264/L` solve `(dA)/(dL) = 0` for L.

`(dA)/(dL) = L - 264/L^2`

`L - 264/L^2 = 0`

=> `L^3 = 264`

=> `L ~~ 6.41`

The width is approximately 3.2 ft.

The width of the box that uses minimum amount of material is approximately 3.2 ft.

Approved by eNotes Editorial Team