#### Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 14

Rs 1000

Hint: For maxima and minima value of A, we must have $\frac{dA}{dl}=0$

Given:  From the question

$l \times b\times 2 =8$

Solution:

Let l,b and h be the length, breadth and height of the tank

h = 2cm

Volume of tank = 8m3

Volume of the tank = $l \times b\times h$

$l \times b\times 2=8$

$lb =4$

$b =\frac{4}{l}$

Area of base $=lb = 4m^3$

Area of 4 walls, $A=2h (l+b)$

\begin{aligned} &\therefore A=4\left(l+\frac{4}{l}\right) \\ &\frac{d A}{d l}=4\left(l-\frac{4}{l^{2}}\right) \\ &\frac{d A}{d l}=0 \\ &4\left(l-\frac{4}{l^{2}}\right)=0 \\ &l=\pm 2 \end{aligned}

The length cannot be negative.

now

\begin{aligned} &\frac{d^{2} A}{d l^{2}}=\frac{32}{l^{3}} \\ &\text { At } l=2 \\ &\frac{d^{2} A}{d l^{2}}=\frac{32}{8}=4>0 \end{aligned}

Thus the area is minimum when l=2,

Cost of the building the base = $Rs 70 \times lb$

$=Rs 70 \times 4$

$=Rs 280$

Cost of the building walls $=Rs\; 2h(l+b)\times 45$

$=Rs\; 90(2)(2+2)$

$=Rs\; 720$

Total cost = Rs (280 +720) = 1000