14. find the maximum and minimum of f(x,y)=x^2 - 4x + y^2+2y in the region g(x,y)=x^2 + y^2≤20. describe…

Explanation:

Step1: Rewrite the function

Rewrite $f(x,y)=x^{2}-4x + y^{2}+2y$ as $f(x,y)=(x - 2)^{2}-4+(y + 1)^{2}-1=(x - 2)^{2}+(y + 1)^{2}-5$.

Step2: Analyze the critical - points

Find the first - order partial derivatives. $\frac{\partial f}{\partial x}=2x-4$, $\frac{\partial f}{\partial y}=2y + 2$. Set $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$.

• From $\frac{\partial f}{\partial x}=0$, we have $2x-4 = 0\Rightarrow x = 2$.
• From $\frac{\partial f}{\partial y}=0$, we have $2y+2 = 0\Rightarrow y=-1$. The critical point is $(2,-1)$. And $f(2,-1)=(2 - 2)^{2}+(-1 + 1)^{2}-5=-5$.

Step3: Use the method of Lagrange multipliers on the boundary $g(x,y)=x^{2}+y^{2}=20$

Let $L(x,y,\lambda)=x^{2}-4x + y^{2}+2y-\lambda(x^{2}+y^{2}-20)$.

• Calculate the partial derivatives: $\frac{\partial L}{\partial x}=2x-4-2\lambda x=0$, $\frac{\partial L}{\partial y}=2y + 2-2\lambda y=0$, $\frac{\partial L}{\partial\lambda}=-(x^{2}+y^{2}-20)=0$.
• From $2x-4-2\lambda x=0$, we get $x - 2-\lambda x=0\Rightarrow x=\frac{2}{1 - \lambda}(\lambda\neq1)$.
• From $2y + 2-2\lambda y=0$, we get $y + 1-\lambda y=0\Rightarrow y=\frac{-1}{1 - \lambda}(\lambda\neq1)$.
• Substitute $x$ and $y$ into $x^{2}+y^{2}=20$: $\left(\frac{2}{1 - \lambda}\right)^{2}+\left(\frac{-1}{1 - \lambda}\right)^{2}=20$.
  • $\frac{4 + 1}{(1 - \lambda)^{2}}=20$, so $(1 - \lambda)^{2}=\frac{1}{4}$, then $1-\lambda=\pm\frac{1}{2}$.
  • Case 1: If $1-\lambda=\frac{1}{2}$, then $\lambda=\frac{1}{2}$, $x = 4$, $y=-2$, and $f(4,-2)=4^{2}-4\times4+(-2)^{2}+2\times(-2)=4 - 4=0$.
  • Case 2: If $1-\lambda=-\frac{1}{2}$, then $\lambda=\frac{3}{2}$, $x=-4$, $y = 2$, and $f(-4,2)=(-4)^{2}-4\times(-4)+2^{2}+2\times2=16 + 16+4 + 4=40$.

Answer:

The maximum value of $f(x,y)$ in the region $x^{2}+y^{2}\leq20$ is $40$ and the minimum value is $-5$.