The function [tex]f(x,y)[/tex] has critical points where the first partial derivatives simultaneously vanish:
[tex]f_x=2x+6=0\implies x=-3[/tex]
[tex]f_y=2y-6=0\implies y=3[/tex]
The point (3, 3) lies within the disk [tex]x^2+y^2\le49[/tex]. The Hessian for [tex]f[/tex] is
[tex]\mathbf H(x,y)=\begin{bmatrix}f_{xx}(x,y)&f_{xy}(x,y)\\f_{yx}(x,y)&f_{yy}(x,y)\end{bmatrix}=\begin{bmatrix}2&0\\0&2\end{bmatrix}[/tex]
which is positive definite for all [tex](x,y)[/tex]. By the second partial derivative test, this indicates the critical point (3, 3) is the site of a minimum, and we have [tex]f(-3,3)=-18[/tex].
Meanwhile, when we consider the boundary, we can set [tex]x=7\cos t[/tex] and [tex]y=7\sin t[/tex], so that [tex]f(x,y)[/tex] can be expressed as a function of one variable [tex]t[/tex]:
[tex]F(t):=f(7\cos t,7\sin t)=49+42(\cos t-\sin t)[/tex]
Then the critical points of [tex]F[/tex] occur at
[tex]F'(t)=-42(\sin t+\cos t)=0\implies\tan t=-1\implies t=-\dfrac\pi4+n\pi[/tex]
where [tex]n[/tex] is any integer. We restrict [tex]0\le t<2\pi[/tex] to omit redundancies, leaving us with just [tex]t=\dfrac{3\pi}4[/tex] and [tex]t=\dfrac{7\pi}4[/tex]. We then have
[tex]F''(t)=-42(\cos t-\sin t)[/tex]
[tex]F''\left(\dfrac{3\pi}4\right)=42\sqrt2>0[/tex]
[tex]F''\left(\dfrac{7\pi}4\right)=-42\sqrt2<0[/tex]
which indicates that a minimum occurs at [tex]t=\dfrac{3\pi}4[/tex], and a maximum at [tex]t=\dfrac{7\pi}4[/tex]. At these points we have values of [tex]F\left(\dfrac{3\pi}4\right)=49-42\sqrt2\approx-10.4[/tex] and [tex]F\left(\dfrac{7\pi}4\right)=49+42\sqrt2\approx108.4[/tex].
So the absolute minimum of [tex]f[/tex] over the disk is -18, and the absolute maximum of [tex]f[/tex] over the disk is [tex]49+42\sqrt2[/tex].
