Question

Assertion ($A$): lf $x+y=12$ then the minimum value of $x^2+y^2$ is 72.
Reason (R): lf $x+y=k$, then the maximum value of xy is $k^2$.

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true and R is not the correct explanation of A
• C. A is true and R is false
• D. A is false and R is true

Answer 1

Answer: (C)

A is true and R is false

$s=x^2+(12-x{)}^2$
$\frac{ds}{dx}=2x+2(12-x)(-1)$
$\frac{ds}{dx}=0$ at $x=6$ $y=6$
$s=x^2+y^2$

$=72$$p=x(k-x)$

$\frac{dp}{dx}=(k-x)-x$
$\frac{dp}{dx}=0$

$\Rightarrow x=\frac{k}{2}$ $y=\frac{k}{2}$

$p=\frac{k^2}{4}$