Answer to Question #107920 in Calculus for annie

Question

Find the maximum and minimum values of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤324.

The maximum value of f(x,y). and List the point(s) where the function attains its maximum as an ordered pair, such as (-6,3), or a list of ordered pairs if there is more than one point, such as (1,3), (-4,7).

Answer 1

$f'_x=4x-4,f_y'=6y$ then the stationary point is $(1,0).$

$f''_{x^2}=4,f''_{xy}=0,f''_{y^2}=6$ then $(1,0)$ is point of the minimum. $f(1,0)=-7$

Examine the $f(x,y)$ at the boundaries of the region

$L(x,y,\lambda)=2x^2+3y^2-4x-5+\lambda(x^2+y^2-324)$

$\lambda'_x=4x-4-2\lambda x,\lambda'_y=6y+2\lambda y$ if $\lambda'_x=\lambda'_y=0$ and $x^2+y^2=324$ then

$\lambda=-3,x=-2,y=8\sqrt5,y=-8\sqrt5$

$\lambda''_{x^2}=4+2\lambda,\lambda''_{xy}=0,\lambda''_{y^2}=6+2\lambda$ hence $d^2\lambda<0$ hence $(-2,8\sqrt5),(-2,-8\sqrt5)$ is points of the conditional maximum. Then $f(-2,8\sqrt5)=f(-2,-8\sqrt5)=971$