Question

Find the maximum and minimum values of
f(x,y) = x+2y on the disk x^2+y^2 ≤1

I have done this for now:

f_1(x,y) = 1
f_2(x,y) = 2

x=cos(t) and y=sin(t)

I have that g(t) = x(t) + 2*y(t) --> g(t) = cost(t) + 2*sin(t)

g'(t) = 0 = 2*cost-sin(t)

Then I can see that:

2cos(t)/cos(t) -sin(t)/cos(t) = 0/cos(t) --> tan = 2

From this point I have no idea what to do.

Answer 1

\[f(x,y)=x+2y\quad,\quad x^2+y^2\leq 1\]\[x=r\cos\phi\quad,\quad y=r\sin\phi\quad;\quad 0\leq r\leq 1\]\[x+2y=r(\cos\phi+2\sin\phi)=\sqrt{5}r\left(\frac{1}{\sqrt{5}}\cos\phi+\frac{2}{\sqrt{5}}\sin\phi\right)\]\[\sin\phi_0=\frac{1}{\sqrt{5}}\quad,\quad\cos\phi_0=\frac{2}{\sqrt{5}}\]\[\Rightarrow f(r,\phi)=\sqrt{5}r\left(\sin\phi_0\cos\phi+\cos\phi_0\sin\phi\right)=\sqrt{5}r\sin{(\phi+\phi_0)}\]\[-\sqrt{5}r\leq\sqrt{5}r\sin{(\phi+\phi_0)}\leq\sqrt{5}r\]\[r=1\quad\Rightarrow\quad f_{\min}=-\sqrt{5}\quad,\quad f_{\max}=\sqrt{5}\]