Chapter 4.7, Problem 16E

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

Chapter
Section

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# A rectangular storage container with an open top is to have a volume of 10 m3. The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs$6 per square meter. Find the cost of materials for the cheapest such container.

To determine

To find: The cost of materials for the cheapest open box container whose volume is 10m3

Explanation

Given:

The volume of the open box container is 10m3.

The length of base is twice the width of the box.

The cost for the base of the box is $10 per square meter. The cost for the sides of the box is$6 per square meter.

Calculation:

Let length is denoted by L, width is denoted by W and height is denoted by H.

Volume of the box =length(L)×width(W)×height(H)

The length is twice the width, L=2W

Therefore, the volume is calculated as follows.

V=L×W×H=10

Substitute the value of L,

2W2×H=10H=102W2

Compute the cost of the materials as follows.

Cost, C=10×Area of the base+6×Area of the sides=10(L×W)+6(2×L×H+2×W×H)=10(2W×W)+6(2×2W×102W2+2×W×102W2)=20W2+120W+60W

Simplifying further,

Cost, C=20W2+180W

Differentiate C with respect to W,

dCdW=ddW(20W2+180W)=20(2W)+180(1W2)=40W180W2

For critical points, set dCdW=0 and solve for W

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