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  • Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 10.

Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 10.

Answers (1)

\frac{25}{4} square units

Hint: For maxima and minima we must have f'(x)=0

Given: x^2 +y^2=5^2

Solution:

Let the base of the triangle be x, and height be y

x^2 +y^2=5^2

y^2=25-x^2

y=\sqrt{25-x^2}

As area of triangle A, A=\frac{1}{2}xy

\begin{aligned} &A=\frac{1}{2} x \times \sqrt{25-x^{2}} \\ &A(x)=\frac{x \sqrt{25-x^{2}}}{2} \\ &A^{\prime}(x)=\frac{\sqrt{25-x^{2}}}{2}+\frac{x(-2 x)}{4 \sqrt{25-x^{2}}} \end{aligned}

\begin{aligned} &=\frac{\sqrt{25-x^{2}}}{2}+\frac{x^{2}}{2 \sqrt{25-x^{2}}} \\ &=\frac{25-x^{2}-x^{2}}{2 \sqrt{25-x^{2}}} \\ &A^{\prime}(x)=\frac{25-2 x^{2}}{2 \sqrt{25-x^{2}}} \\ &A^{\prime}(x)=0 \end{aligned}

\begin{aligned} &\frac{25-2 x^{2}}{2 \sqrt{25-x^{2}}}=0 \\ &25-2 x^{2}=0 \\ &x=\frac{5}{\sqrt{2}}, y=\sqrt{25-\frac{25}{2}} \\ &x=\frac{5}{\sqrt{2}}, y=\frac{5}{\sqrt{2}} \end{aligned}

Also A''(x) =\frac{\left[-4 x \sqrt{25-x^{2}}-\frac{\left(25-2 x^{2}\right)(-2 x)}{2 \sqrt{25-x^{2}}}\right]}{25-x^{2}}

\begin{aligned} &=\frac{\left[\frac{-4 x\left(25-x^{2}\right)+\left(25-2 x^{3}\right)}{2 \sqrt{25-x^{2}}}\right]}{25-x^{2}} \\ &=\frac{-100 x+4 x^{3}+25 x-2 x^{3}}{\left(25-x^{2}\right) \sqrt{25-x^{2}}} \\ &=\frac{-75 x+2 x^{3}}{\left(25-x^{2}\right) \sqrt{25-x^{2}}} \end{aligned}

A^{\prime \prime}\left(\frac{5}{\sqrt{2}}\right)=\frac{-75\left(\frac{5}{\sqrt{2}}\right)+2\left(\frac{5}{\sqrt{2}}\right)^{3}}{\left(25-\left(\frac{5}{\sqrt{2}}\right)^{2}\right)^{\frac{3}{2}}}<0

So, x=\left (\frac{5}{\sqrt{2}} \right ) is point of maxima

\therefore Largest possible area of triangle,

=\frac{1}{2}\left(\frac{5}{\sqrt{2}}\right)\left(\frac{5}{\sqrt{2}}\right)=\frac{25}{4} square units

Posted by

infoexpert24

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