Optimization/ Extreme Value Problem

Question

you are an engineer for Jiffy box Company. it is your job to design an open topped( that is, no top) cardboard box with a square base. The box must have a volume of 134 cubic inches of space available for storage. use calculus to determine the dimensions of the box that will minimize surface area and subsequently minimize cost of the company. Show all work! Round final answers to the nearest hundredth.

(Note: Vbox = LWH; SA open—top box = LW + 2LH + 2WH)

Answers needed: length in inches, width in inches, and heigh in inches.

Answer 1

A square base means length = width so let's let both length and width = "x"

Then, because the volume is 134, we can say 134 = x²•h or h = 134/x²

Then, because SA = LW + 2LH + 2WH

we can say:

SA = x² + 2x•h + 2x•h

SA = x² + 4x•h

SA(x) = x² + 4x•(134/x²)

SA(x) = x² + 536/x

To minimize SA, take the derivative and set it equal to zero:

SA'(x) = 2x - 536/x²

2x - 536/x² = 0

2x = 536/x²

x³ = 268

x = 6.447 inches

Using our equation h = 134/x²

h = 134/(6.447)²

h = 3.224 inches

So:

Length: 6.45 inches

Width: 6.45 inches

Height: 3.22 inches

Answer 2

Andrea

Draw a picture.

Let's call the length of the box x.

Then the width must also be x as base is a square.

As volume is 134, height must be 134/x²

Surface area A = area of bottom + 4 x area of vertical sides

A = x² + 4 . (134/x²) = x² + (536/x²)

To minimise A , you need to solve dA/dx = 0. Exercise for you.

Mike