Question

find the maximum and minimum value of xy(a-x-y)
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Answer 1

Answer

The equation of the curve is :

A = x y(a-x-y)

A=a x y -x²y-x y²

Calculating the partial derivatives of the above function

$\frac{\partial A }{\partial x}=a y-2 x y-y^2\\\\\frac{\partial A}{\partial y}=ax-x^2-2 x y\\\\\frac{\partial^2A }{\partial x^2}=-2 y\\\\\frac{\partial^2A }{\partial y^2}=-2x$

$\frac{\partial^2A }{\partial x\partial y}=a-2 x-2 y$

Calculating the critical points

$a y-2 x y-y^2=0\\\\y(a -2 x-y)=0\\\\y=0\\\\2 x+y=a\\\\ax-x^2-2 x y=0\\\\x(a-x-2 y)=0\\\\x=0\\\\x+2 y=a$

2 x+y=a-----(1)

x+2 y=a------(2)

Equation (1) - 2 ×equation (2)

2 x+y-2 x-4 y=a-2 a

-3 y =-a

$y=\frac{a}{3}\\\\ x+\frac{2a}{3}=a\\\\x=a-\frac{2a}{3}\\\\x=\frac{a}{3}$

So, critical points are , (0,0) and $(\frac{a}{3},\frac{a}{3})$

$\frac{\partial^2 A}{\partial x^2 }(_{0,0})=0$

$\frac{\partial^2 A}{\partial x^2}(_{\frac{a}{3},\frac{a}{3}})=\frac{-2 a}{3}$

$\Delta=\frac{\partial^2A }{\partial x^2}\frac{\partial^2A }{\partial x^2}-\frac{\partial^2 A}{\partial x \partial y}\\\\ \Delta(_{0,0})=0\times 0-a=-a\\\\\Delta (_{\frac{a}{3},\frac{a}{3}})=\frac{-2a}{3}*\frac{-2a}{3}+\frac{a}{3}=\frac{4a^2}{9}+\frac{a}{3}$

$\Delta(_{0,0})>0,\frac{\partial^2A }{\partial x^2}>0,$

So, (0,0) point of Minima.

Minimum value =0

$\Delta (_{\frac{a}{3},\frac{a}{3}})>0, \frac{\partial^2A }{\partial x^2}<0$

So, $(\frac{a}{3},\frac{a}{3})$ is point of Maxima.

Maximum value

$=\frac{a}{3}*\frac{a}{3}(a-\frac{a}{3}-\frac{a}{3})\\\\=\frac{a^2}{9}*\frac{a}{3}\\\\=\frac{a^3}{27}$