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Maxima and Minima — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMaxima and Minima5 Marks
Question
An open tank is to be constructed with a square base and vertical sides so as to contain a given quantity of water. Show that the expenses of lining with lead with be least, if depth is made half of width.
✓

Answer

Let $x$ be side of the sqqre base and $h$ be the depth of the given tank of volume $V.$
$\therefore V = x^2h$
$\Rightarrow \text{h}=\frac{\text{V}}{\text{x}^{2}} \ ...(\text{i})$
Surface area of the cost per sqqre unit of material and $C$ be the total cost.
$\text{C}=(\text{x}^{2}+4\text{xh})\text{p}=\Big(\text{x}^{2}+\frac{4\text{V}}{\text{x}}\Big)\text{p}$.
Where $p$ is constant'
$\frac{\text{dC}}{\text{dx}}=\text{p}\Big(2\text{x}-\frac{4\text{V}}{\text{x}^{2}}\Big)$
Now, $\frac{\text{dC}}{\text{dx}}=0$ gives us,
$=\text{p}\Big(2\text{x}-\frac{4\text{V}}{\text{x}^{2}}\Big)=0$ or $2\text{x}-\frac{4\text{V}}{\text{x}^{2}}=0$
$\therefore x^{_3} - 2V = 0$
$\Rightarrow x^3 = 2V$
$\frac{\text{d}^{2}\text{C}}{\text{dx}^{2}}=\text{p}\Big(2+\frac{8\text{V}}{\text{x}^{2}}\Big)=\text{p}\Big(2+\frac{8}{\text{x}^{3}}.\frac{\text{x}^{3}}{2}\Big)$
$=p(2 + 4) = 6p > 0$
Now, $\text{h}=\frac{\text{V}}{\text{x}^{2}}=\frac{\frac{\text{x}^{3}}{2}}{\text{x}^{2}}=\frac{\text{x}}{2}$
$\therefore$ depth of tank $=$ half of its width.

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