Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Arectangular 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 material for the cheapest such container.

The cost of materials is $100.4$ dollars.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Data

1)Volume of the container is 10 m³.

2)The length of the base is twice the width.

3) Material for the base costs $10 per square meter.

4) Material for the sides costs $6 per square meter.

Step 2: Determination of the cost of material

Letx be the width of box and 2x be length of the base.

The base area, $A = x \cdot 2x = 2{x^2}$

Since the volume is$10\;{{\mathop{\rm m}\nolimits} ^3}$, the height will be $h = \frac{{10}}{{2{x^2}}} = \frac{5}{{{x^2}}}$

The cost of making such a container is

Cost of base$ = 2{x^2} \times 10 = 20{x^2}$

Cost of sides=$\left( {\left( {2 \cdot 2x \cdot \frac{5}{{{x^2}}}} \right) + \left( {2 \cdot x \cdot \frac{5}{{{x^2}}}} \right)} \right) \times 6$

$ = \frac{{80}}{x}$

The total cost = cost of base + cost of sides.

$f\left( x \right) = 20{x^2} + \frac{{80}}{x}$

$f\left( x \right) = 20\left( {{x^2} + \frac{4}{x}} \right)$

Find first derivative of$f\left( x \right)$ and equate to 0 to get the minimum or cheapest cost.

$f\left( x \right) = 20\left( {{x^2} + \frac{4}{x}} \right)$

$f'\left( x \right) = 20\left( {2x - \frac{4}{{{x^2}}}} \right)$

$\begin{aligned}{c}20\left( {2x - \frac{4}{{{x^2}}}} \right) &= 0\\2x &= \frac{4}{{{x^2}}}\\2{x^3} &= 4\\{x^3} &= 4\\x &= 1.58\end{aligned}$

Thus, $x = 1.58\;{\mathop{\rm m}\nolimits} $.

$f\left( x \right) = 20\left( {{x^2} + \frac{4}{x}} \right)$

Put $x = 1.58\;{\mathop{\rm m}\nolimits} $into $f\left( x \right) = 20\left( {{x^2} + \frac{4}{x}} \right)$in above equation.

$\begin{aligned}{c}f\left( {1.58} \right) &= 20\left( {{{1.58}^2} + \frac{4}{{1.58}}} \right)\\ &= 100.4\;{\rm{dollars}}\end{aligned}$

Thus, the cost of materials is $100.4$ dollars.

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Essential Calculus: Early Transcendentals

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Short Answer

Arectangular 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 material for the cheapest such container.\(\)

Step by Step Solution

Step 1: Given Data

Step 2: Determination of the cost of material

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Arectangular 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 material for the cheapest such container.\(\)

Step by Step Solution

Step 1: Given Data

Step 2: Determination of the cost of material

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Arectangular 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 material for the cheapest such container.\(\)