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

Answers (1)

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


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.


\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

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