Question

Tìm GTLN

A=-x² -x-y² -3y +13

Answer 1

A=\(-x^2-x-y^2-3y+13=-\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}\right)-\left(y^2+3.y.\frac{1}{9}\right)+13+\frac{1}{4}+\frac{1}{9}\)

= \(-\left(x+\frac{1}{2}\right)^2-\left(y+\frac{1}{3}\right)^3+\frac{481}{26}\)

ta có : \(\left(x+\frac{1}{2}\right)^2\ge0\)=> \(-\left(x+\frac{1}{2}\right)^2\le0\)

               \(\left(y+\frac{1}{3}\right)^2\ge0\)=> \(-\left(y+\frac{1}{3}\right)^2\le0\)

=> \(-\left(x+\frac{1}{2}\right)^2-\left(y+\frac{1}{3}\right)^3+\frac{481}{26}\)\(\le\frac{481}{36}\)

=> Max_A=\(\frac{481}{36}\)khi \(\begin{cases}x+\frac{1}{2}=0\\y+\frac{1}{3}=0\end{cases}\)<=> x=-1/2 và y=-1/3