#### Answer

$39$

#### Work Step by Step

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, $f(x,y)=49-x^2-y^2 =49-1-9=39$