Question

An open metal tank of square base has a volume of 123 m^3

Answer 1

Answer:

  a) h = 123/x^2

  b) S = x^2 +492/x

  c) x ≈ 6.27

  d) S'' = 6; area is a minimum (Y)

  e) Amin ≈ 117.78 m²

Step-by-step explanation:

a) The volume is given by ...

  V = Bh

where B is the area of the base, x^2, and h is the height. Filling in the given volume, and solving for the height, we get:

  123 = x^2·h

  h = 123/x^2

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b) The surface area is the sum of the area of the base (x^2) and the lateral area, which is the product of the height and the perimeter of the base.

$S=x^2+Ph=x^2+(4x)\dfrac{123}{x^2}\\\\S=x^2+\dfrac{492}{x}$

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c) The derivative of the area with respect to x is ...

$S'=2x-\dfrac{492}{x^2}$

When this is zero, area is at an extreme.

$0=2x -\dfrac{492}{x^2}\\\\0=x^3-246\\\\x=\sqrt[3]{246}\approx 6.26583$

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d) The second derivative is ...

$S''=2+\dfrac{2\cdot 492}{x^3}=2+\dfrac{2\cdot 492}{246}=6$

This is positive, so the value of x found represents a minimum of the area function.

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e) The minimum area is ...

$S=x^2+\dfrac{2\cdot 246}{x}=(246^{\frac{1}{3}})^2+2\dfrac{246}{246^{\frac{1}{3}}}=3\cdot 246^{\frac{2}{3}}\approx 117.78$

The minimum area of metal used is about 117.78 m².