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Please Solve RD Sharma Class 12 Chapter 17 Maxima and Minima Exercise Case Study Based Questions question 5 subquestion (iv) Maths Textbook Solution.

Answers (1)

Answer: $x = 2$

Hint: use the concept of maxima and minima.

Solution:

As we know,

$C^{\prime}(x)=3600\left(1-\frac{4}{x^{2}}\right)$

Putting $c^{\prime}(x)=0$

$\begin{aligned} 3600\left(1-\frac{4}{x^{2}}\right) &=0 . \\ \end{aligned}$

$1-\frac{4}{x^{2}} =0 \\$

$\frac{x^{2}-4}{x^{2}} =0 \\$

$x^{2}-4=0 \\$

$(x-2)(x+2) =0$

$\begin{aligned} &\mathrm{x}=2 \\ \end{aligned}$

$\mathrm{c}^{\prime \prime}(\mathrm{x})=3600\left(0-4(-2) \mathrm{x}^{-2-1}\right) \\$

$\quad=3600(0-4(-2) \mathrm{x}^{-3} \\$

$=3600\left(8 \mathrm{x}^{-3}\right) \\$

$=28800 \mathrm{x}^{-3} \\$

$=\frac{28800}{\mathrm{x}^{3}}$

Putting $\begin{aligned} &\mathrm{x}=2 \\ \end{aligned}$

$\begin{aligned} &c^{\prime \prime}(2)=\frac{28800>0}{(2)^{3}} \\ &c^{\prime \prime}(2)>0 \end{aligned}$

So minimum is $\begin{aligned} &\mathrm{x}=2 \\ \end{aligned}$

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