   Chapter 3.7, Problem 16E

Chapter
Section
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 such container

Explanation

1) Concept:

i. First derivative test for absolute extreme values- suppose that c is a critical number of a continuous function f defined on an interval.

a) If f'x>0 for all x<c and f'x<0 for all x>c, then fc is the absolute maximum value of f

b) If f'x<0 for all x<c and f'x>0 for all x>c, then fc is the absolute Minimum value of f

ii. The cost is 10 times the area of the base (L·W) plus 6 times the total area of the sides.

iii. A critical number of a function f   is a number c in the domain of f  such that either  f'c=0 or f'c does not exist.

2) Formula:

Volume of the rectangle =l·h·w

3) Given:

The rectangular container has volume 10m3 ; also its length is twice the width, and the cost is $10 per square meter of the base and$6 per square meter of the sides.

4) Calculation:

Let l be length, w be width and h be height of a rectangle

So volume of rectangle is V=l·w·h

Given that V=10 and l=2·w

Now put l=2·w in the equation l·w·h=10

(2·w)·w·h=10

2w2·h=10

Divide by 2w2

2w2·h2w2=102w2

h=102w2

Now we have to find cost

C =10l·w+6(2·l·h+2·w·h)

Put l=2·w, h=102w2  in the above equation, =102·w·w+62·2·w·102w2+2·w·102w2

Simplifying,

=102w2+64w·102w2+2·w·102w2

=20</

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