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  • Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 7.

Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 7.

Answers (1)

\frac{112}{\pi +4}\; \text{and}\; \frac{28 \pi }{\pi +4}

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

Given: Suppose the wire which is to be made into a square and circle is cut into two pieces of length x,m and ym respectively.

Solution: x +y = 28                  ......(1)

Perimeter of square 4a=x

a=\frac{x}{4}

Area of square a^2 =\left ( \frac{x}{4} \right )^4=\frac{x^2}{16}

Circumference of circle 2\pi r=y

r=\frac{y}{2\pi}

Area of circle =\pi r^2

=\pi \left ( \frac{y}{2\pi } \right )^2=\frac{y^2}{4\pi }

z = Area of square + Area of circle

\begin{aligned} &=\frac{x^{2}}{16}+\frac{y^{2}}{4 \pi} \\ &=\frac{x^{2}}{16}+\frac{(28-x)^{2}}{4 \pi} \\ &\frac{d z}{d x}=\frac{2 x}{16}-\frac{2(28-x)}{4 \pi} \end{aligned}

For maximum or minimum value of z 

\frac{dz}{dx}=0

\begin{aligned} &\frac{2 x}{16}-\frac{2(28-x)}{4 \pi}=0 \\ &\frac{x}{4}=\frac{(28-x)}{\pi} \\ &\frac{x \pi}{4}+x=28, x\left(\frac{\pi}{4}+1\right)=28 \end{aligned}

\begin{aligned} x &=\frac{28}{\left(\frac{\pi}{4}+1\right)} \\ x &=\frac{112}{\pi+4} \end{aligned}

y=28-\frac{112}{\pi +4}                from equation 1

\frac{d^{2} z}{d x^{2}}=\frac{1}{8}+\frac{1}{2 \pi}>0

Thus z is maximum when \frac{112}{\pi +4} and \frac{28 \pi }{\pi +4}

Posted by

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