If x+y=6, x geqslant 0, y geqslant 0,then the | vedclass

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(B) Let $f(x, y) = x^2 y$. Given $x+y=6$,we have $y = 6-x$.Substituting $y$ in $f$,we get $f(x) = x^2(6-x) = 6x^2 - x^3$.To find the maximum,we find t

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) Let $f(x, y) = x^2 y$. Given $x+y=6$,we have $y = 6-x$.Substituting $y$ in $f$,we get $f(x) = x^2(6-x) = 6x^2 - x^3$.To find the maximum,we find the derivative $f'(x) = 12x - 3x^2$.Setting $f'(x) = 0$,we get $3x(4-x) = 0$,which gives $x=0$ or $x=4$.Since $x=0$ gives $f(0)=0$,we check $x=4$.For $x=4$,$y = 6-4 = 2$.The value is $f(4) = 4^2 	imes 2 = 16 	imes 2 = 32$.Thus,the maximum value is $32$.

If $x+y=6, x \geqslant 0, y \geqslant 0$,then the maximum value of $x^2 y$ is

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