Question

Find the local maxima, local minima, and saddle points, if any, for the function z = 3x3 – 36xy – 3y3. (Use symbolic notation and fractions where needed. Give your answer as point coordinates in t

Answer 1

(0,0) is a saddle point

(-4,4) is a local maximum

Explanation:

`\displaystyle z=3x^3-36xy-3y^3\\\\(\partial z)/(\partial x)=9x^2-36y\\\\(\partial z)/(\partial y)=-36x-9y^2`

Determine critical points

`9x^2-36y=0\\9x^2=36y\\(x^2)/(4)=y`

`-36x-9y^2=0\\-36x-9((x^2)/(4))^2=0\\-36x-(9)/(16)x^4=0\\x(-36-(9)/(16)x^3)=0\\\\x=0\\\\-36-(9)/(16)x^3=0\\-36=(9)/(16)x^3\\-64=x^3\\-4=x`

When x=0

`9x^2-36y=0\\9(0)^2-36y=0\\-36y=0\\y=0`

When x=-4

`9x^2-36y=0\\9(-4)^2-36y=0\\9(16)-36y=0\\144-36y=0\\144=36y\\4=y`

So, we need to check what kinds of points (0,0) and (-4,4) are.

For (0,0)

`\displaystyle H=\biggr((\partial^2 z)/(\partial x^2)\biggr)\biggr((\partial^2 z)/(\partial y^2)\biggr)-\biggr((\partial^2 z)/(\partial x\partial y)\biggr)^2\\\\H=(18x)(-18y)-(-36)^2\\\\H=(18(0))(-18(0))-(-36)^2\\\\H=-1296 < 0`

Therefore, (0,0) is a saddle point since
`H < 0`.

For (-4,4)

`\displaystyle H=\biggr((\partial^2 z)/(\partial x^2)\biggr)\biggr((\partial^2 z)/(\partial y^2)\biggr)-\biggr((\partial^2 z)/(\partial x\partial y)\biggr)^2\\\\H=(18x)(-18y)-(-36)^2\\\\H=(18(-4))(-18(4))-(-36)^2\\\\H=(-72)(-72)-1296\\\\H=5184-1296\\\\H=3888 > 0`

Because
`H > 0` and since
`(\partial^2z)/(\partial x^2)=-72 < 0`, then (-4,4) is a local maximum