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

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

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

Posted by

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