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Application of Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives5 Marks
Question
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is $8\ m^3.$ If building of tank costs Rs $70$ per sq.metres for the base and Rs $45$ per sq. metre for sides. What is the cost of least expensive tank ?
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Answer

Let $x$ and $y$ be the length and width of rectangular base, $v$ be the volume.$v = 8 ($Given$)$
$v = 2xy$
$8 = 2xy$
$y = \frac{4}{x}$
$s = (xy) \times 70 + 2(x + y) \times 45$
$ = x \times \frac{4}{x} \times 70 + 90\left( {x + \frac{4}{x}} \right)$
$ = 280 + 90\left( {x + \frac{4}{x}} \right)$
$\frac{{ds}}{{dx}} = 0 + 90\left( {1 - \frac{4}{{{x^2}}}} \right)$
=$\frac{{{d^2}s}}{{d{x^2}}} = 90\left( {0 + \frac{8}{{{x^3}}}} \right)$
For maximum/minimum
$\frac{{ds}}{{dx}} = 0$
$x = 2$
${\left( {\frac{{{d^2}s}}{{d{x^2}}}} \right)_{x = 2}}=\frac{{720}}{{{2^3}}} > 0$
s is Minimum at $x=2$
Minimum cost is given by
$s = 280 + 90\left( {2 + \frac{4}{2}} \right)$
$= 280+ 90 (4)$
$= 280+360$
$= 640$

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