Thomas' Calculus 13th Edition

$39$
Given: $f(x,y) =49-x^2-y^2$ $g(x,y)=x+3y-10=0$ ...(1) Consider the gradient equation as follows: $\nabla =\lambda \nabla g$ Here, $-2x=\lambda$ ; $-2y=3 \lambda$ This implies that $x=-\dfrac{ \lambda}{2}; y=-\dfrac{3\lambda}{2}$ Plugging $x=-\dfrac{ \lambda}{2}; y=-\dfrac{3\lambda}{2}$ in equation (1), we get $g(x,y)=x+3y-10=0$ This implies that $\lambda=-2$ and $x=1 ; y=3$ Hence, $f(x,y)=49-x^2-y^2 =49-1-9=39$