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

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

Answers (1)

A=r^2

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)

Now,

\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

Posted by

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