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Q14E

Expert-verified
Found in: Page 238

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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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.

The cost of materials is $$100.4$$ dollars.

See the step by step solution

## 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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