Question 858571
{{{L=x^2+y^2}}}
Substitute,
{{{x-y=12}}}
{{{x=12+y}}}
.
.
{{{L=(12+y)^2+y^2}}}
{{{L=(144+24y+y^2)+y^2}}}
{{{L=2y^2+24y+144}}}
Take the derivative of L with respect to y and set it to zero.
{{{dL/dy=4y+24=0}}}
{{{4y+24=0}}}
{{{4y=-24}}}
{{{y=-6}}}
Then,
{{{x=12-6}}}
{{{x=6}}}