First find the critical points of f :
[tex]f(x,y)=2x^2+3y^2-4x-5=2(x-1)^2+3y^2-7[/tex]
[tex]\dfrac{\partial f}{\partial x}=2(x-1)=0\implies x=1[/tex]
[tex]\dfrac{\partial f}{\partial y}=6y=0\implies y=0[/tex]
so the point (1, 0) is the only critical point, at which we have
[tex]f(1,0)=-7[/tex]
Next check for critical points along the boundary, which can be found by converting to polar coordinates:
[tex]f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t[/tex]
Find the critical points of g :
[tex]\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0[/tex]
[tex]\implies\sin t=0\text{ OR }1+5\cos t=0[/tex]
[tex]\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi[/tex]
where n is any integer. We get 4 critical points in the interval [0, 2π) at
[tex]t=0\implies f(10,0)=155[/tex]
[tex]t=\cos^{-1}\left(-\dfrac15\right)\implies f(-2,4\sqrt6)=299[/tex]
[tex]t=\pi\implies f(-10,0)=235[/tex]
[tex]t=2\pi-\cos^{-1}\left(-\dfrac15\right)\implies f(-2,-4\sqrt6)=299[/tex]
So f has a minimum of -7 and a maximum of 299.
