A tank with rectangular base and rectangular sides, open at the t… - Vidyadip

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 square metre for sides. What is the cost of least expensive tank?

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Answer

Let $l,$ and hrepresent the length, breadth, and height of the tank respectively.
Then, we have height $(h) = 2\ m$
Volume of the tank $= 8\ m^3$
Volume of the tank = $\text{l}\times\text{b}\times\text{h}$
$\therefore\ 8=\text{l}\times\text{b}\times2$
$\Rightarrow\ \text{lb}=4\Rightarrow\ \text{b}\frac{4}{\text{l}}$
Now, area of the base $= lb = 4$
Area of the $4$ Walls $ (A) - 2h (l + b)$
$\therefore\ \text{A}=4\Big(\text{l}+\frac{4}{\text{l}}\Big)$
$\Rightarrow\frac{\text{dA}}{\text{dl}}=4\Big(1-\frac{4}{\text{l}^2}\Big)$
Now, $\frac{\text{dA}}{\text{dl}}=0$
$\Rightarrow\ 1-\frac{4}{\text{l}^2}=0$
$\Rightarrow\ \text{l}^2=4$
$\Rightarrow\ \text{l}=\pm2$
However, the length cannot be negative.
Therefore, we have $l = 4.$
$\therefore\ \text{b}=\frac{4}{\text{l}}=\frac{4}{2}=2$
Now, $\frac{\text{d}^2\text{A}}{\text{dl}^2}=\frac{32}{\text{l}^3}$
When l - 2, $\frac{\text{d}^2\text{A}}{\text{dl}^2}=\frac{32}{8}=44>0.$
Thus, by second derivative test, the ares is the minimum when $l = 2.$
we have $l = b = h = 2.$
$\therefore$ Cost of building the base $= Rs. 70 \times (Lb) = Rs. 70(4) = Rs. 280$
Cost of building the walls $= Rs. 2h(L + b) \times 45 = Rs. 90(2)(2 + 2) = Rs. 8(90) = Rs. 720$
Required total cost $= Rs. (280 + 720) = Rs. 1000$
Hence, the tatal cost of the tank will be $Rs. 1000.$

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