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  • Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 3 maths textbook solution.

Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 3 maths textbook solution.

Answers (1)

\left (\frac{1}{2}-\frac{1}{\sqrt{3}} \right )\text{ and} \; \frac{1}{\sqrt{3}}

Hint: For maximum or minimum value of z must have \frac{dz}{dx}=0

Given: x,y >-2 and x+y =\frac{1}{2}

Solution: Let the numbers be x and y then

x,y >-2 and x+y =\frac{1}{2}                             .....(1)

\begin{aligned} &\text { Now, }\\ &z=x+y^{3}\\ &z=x+\left(\frac{1}{2}-x\right)^{3} \quad[\text { from }(1)]\\ &\frac{d z}{d x}=1+3\left(\frac{1}{2}-x\right)^{2} \end{aligned}

For maximum or minimum value of z

\frac{dz}{dx}=0

\begin{aligned} &1+3\left(\frac{1}{2}-x\right)^{2}=0 \\ &\left(\frac{1}{2}-x\right)^{2}=\frac{1}{3} \\ &\left(\frac{1}{2}-x\right)=\pm \frac{1}{\sqrt{3}} \\ &x=\frac{1}{2} \pm \frac{1}{\sqrt{3}} \end{aligned}

\begin{aligned} &\frac{d^{2} z}{d x^{2}}=3-6\left(\frac{1}{2}+\frac{1}{\sqrt{3}}\right) \\ &=\frac{-6}{\sqrt{3}}<0 \end{aligned}

\therefore z is minimum when x=\frac{1}{2}+\frac{1}{\sqrt{3}}

At x=\frac{1}{2}-\frac{1}{\sqrt{3}}

\begin{aligned} &\frac{d^{2} z}{d x^{2}}=3-6\left(\frac{1}{2}-\frac{1}{\sqrt{3}}\right) \\ &=\frac{-6}{\sqrt{3}}>0 \end{aligned}

\therefore z is minimum when x=\frac{1}{2}-\frac{1}{\sqrt{3}}

x+y = \frac{1}{2}

Substituting the value in (1)

\begin{aligned} &y=\frac{-1}{2}+\frac{1}{\sqrt{3}}+\frac{1}{2} \\ &y=\frac{1}{\sqrt{3}} \end{aligned}

So the two required numbers are \left (\frac{1}{2}-\frac{1}{\sqrt{3}} \right )\text{ and} \; \frac{1}{\sqrt{3}}

 

 

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