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Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 1 maths textbook solution.

Answers (1)

$x= \frac{15}{2}, y =\frac{15}{2}$

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

Given: $x+y =15$

Solution: $x+y =15$ ..........(1)

Now,

$z= x^2+y^2$

$z= x^2+(15-x)^2$ (from equation 1)

$z= x^2+x^2+225 -30x$ $[(a-b)^2=a^2+b^2-2ab]$

$\frac{dz}{dx}=4x-30$

For maximum or minimum value of z

$\frac{dz}{dx}=0$

$4x-30=0$

$x=\frac{15}{2}$

$\frac{d^2z}{dx^2}=4>0$

Substituting $x= \frac{15}{2}$ in (1)

$y= \frac{15}{2}$

$\therefore$ z is minimum when x = y = $\frac{15}{2}$

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