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Q14E

Expert-verifiedFound in: Page 238

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**Arectangular storage container with an open top is to have a volume of 10 m ^{3}. 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.

1)Volume of the container is 10 m^{3}.

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.

Let*x* be the width of box and 2*x* 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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