Question

Find the extreme values of f subject to both constraints f(x,y) = 2x^2+3y^2-4x-5, x^2 + y^2 <=16

Answer 1

`f(x,y)=2x^2+3y^2-4x-5`

`f_x=4x-4=0\implies x=1`

`f_y=6y=0\implies y=0`


`f(x,y)` has only one critical point at
`(x,y)=(1,0)`. The function has Hessian


`\mathbf H(x,y)=\begin{bmatrix}f_(xx)&f_(xy)\\f_(yx)&f_(yy)\end{bmatrix}=\begin{bmatrix}4&0\\0&6\end{bmatrix}`

which is positive definite for all
`(x,y)`, which means
`f(x,y)` attains a minimum at the critical point with a value of
`f(1,0)=-7`.

To find the extrema (if any) along the boundary, parameterize it by
`x=4\cos t` and
`y=4\sin t`, with
`0\le t<2\pi`. On the boundary, we have


`f(x(t),y(t))=F(t)=2(4\cos t)^2+3(4\sin t)^2-4(4\cos t)-5=32\cos^2t+48\sin^2t-16\cos t-5`

`F(t)=35-16\cos t-8\cos2t`

Find the critical points along the boundary:


`F'(t)=16\sin t+16\sin2t=16\sin t+32\sin t\cos t=16\sin t(1+2\cos t)=0`

`\implies t=0,\frac{2\pi}3,\pi,\frac{4\pi}3`

Respectively, plugging these values into
`F(t)` gives 11, 47, 43, and 47. We omit the first and third, as we can see the absolute extrema occur when
`F(t)=47`.

Now, solve for
`x,y` for both cases:


`t=\frac{2\pi}3\implies\begin{cases}x=4\cos t=-2\\y=4\sin t=2\sqrt3\end{cases}`


`t=\frac{4\pi}3\implies\begin{cases}x=4\cos t=-2\\y=4\sin t=-2\sqrt3\end{cases}`

so
`f(x,y)` has two absolute maxima at
`(x,y)=(-2,\pm2\sqrt3)` with the same value of 47.