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Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 18.

Answers (1)


Hint: For maxima or minima f'(x) must be zero f'(x) = 0

Given:  \frac{x^2}{4}+y^2=r^2

Solution: \frac{x^2}{4}+y^2=r^2

x^2 +4y^2=4r^2

x^2 =4(r^2 -y^2)                ......(1)

Area = xy

Square on both sides

A^2=x^2 y^2

Z= 4y^2(r^2-y^2)                ......from (1)


\begin{aligned} &\frac{d Z}{d y}=8 y r^{2}-16 y^{3} \\ &\frac{d Z}{d y}=0 \\ &8 y r^{2}-16 y^{3}=0 \\ &8 r^{2}=16 y^{2} \\ &y^{2}=\frac{r^{2}}{2}, y=\frac{r}{\sqrt{2}} \end{aligned}

Substitute y value in eqn (1)

\begin{aligned} &x^{2}=4\left(r^{2}-\left(\frac{r}{\sqrt{2}}\right)^{2}\right) \\ &=4\left(r^{2}-\frac{r^{2}}{2}\right) \\ &=4\left(\frac{r^{2}}{2}\right) \end{aligned}

x^2 = 2r^2

x= r\sqrt{2}

\begin{aligned} &\frac{d^{2} Z}{d y^{2}}=8 r^{2}-48 y^{2} \\ &\frac{d^{2} Z}{d y^{2}}=8 r^{2}-48\left(\frac{r^{2}}{2}\right) \\ &=-16 r^{2}<0 \end{aligned}

Thus the area is maximum when x=r\sqrt{2} and y= \frac{r}{\sqrt{2}}

Area = xy

=r \sqrt{2} \times \frac{r}{\sqrt{2}}

A=r ^2

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