   Chapter 3.7, Problem 17E

Chapter
Section
Textbook Problem

# Do Exercise 16 assuming the container has a fid that is made from the same material as the sides.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 (x·y) 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 s uch that either  f'c=0 or f'c does not exist.

2) Formula:

Volume of the rectangle =l·w·h

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 and the lid

4) Calculation:

Let x be the length, y be the width and z be the height of a rectangle

So, volume of the rectangle is V=x·y·z

Given that V=x·y·z=10 and x=2·y

Now put x=2·y in the equation x·y·z=10

(2·y)·y·z=10

2y2·z=10

Divide by 2y2

2y2·z2y2=102y2

z=102y2

Now we have to find the cost

C=10x·y+6(2·x·z+2·y·z)

Put x=2·y, z=102y2  in the above equation  =102·y·y+62·2·y·102y2+2·y·102y2+6(2x2)

On simplifying,

=102y2+64y·102y2+2·y·102y2+6(2x2)

=32y2+240y<

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