91Ó°ÊÓ

The given function is $f(x,y)=x^2+y$and the constraint is$x^2+y^2≤4.$

To find the extreme of the function we will use Lagrange's method.

Let's find the gradient of the given functions,

$$\begin{gathered} \mathrm{∇}f(x,y)=2x\mathbf{i}+\mathbf{j} \\ \mathrm{∇}\mathrm{g}(\mathrm{x},\mathrm{y},\mathrm{z})=2\mathrm{x}\mathbf{i}+2\mathrm{y}\mathbf{j} \end{gathered}$$

So, the system of equations is

$$\begin{gathered} \mathrm{∇}f(x,y)=\mathrm{λ}\mathrm{∇}g(x,y) \\ 2x\mathbf{i}+\mathbf{j}=\mathrm{λ}(2x\mathbf{i}+2y\mathbf{j}) \end{gathered}$$

Here,$2x=2\mathrm{λ}x,\text{and}1=2\mathrm{λ}y.$

The values of $\mathrm{λ},$we get are$\mathrm{λ}=\frac{2x}{2x}=\frac{1}{2y}.$

So,

role="math" localid="1649944468377" $$\begin{gathered} 4xy=2x \\ 2x(2y−1)=0 \\ x=0\mathrm{and}2y-1=0 \\ y=\frac{1}{2} \end{gathered}$$

Substitute the above values in the given constraint,

$$\begin{gathered} \mathrm{When}\mathrm{x}=0, \\ {x}^2+y^2=4 \\ y=Â\pm 2 \\ \mathrm{When}y=\frac{1}{2}, \\ {x}^2+y^2=4 \\ {x}^2=4-\frac{1}{4} \\ x=Â\pm \frac{\sqrt{15}}{2} \end{gathered}$$

Thus, the points where the extreme value appears are$(0,−2),(0,2),\left(−\frac{\sqrt{15}}{2},\frac{1}{2}\right),\left(\frac{\sqrt{15}}{2},\frac{1}{2}\right).$

Now, let's find the value of the function through the points.

So,

When the point is $\left(0,-2\right)$the value of the function is $-2.$

When the point is $\left(0,2\right)$the value of the function is $2.$

And when the points are$\left(−\frac{\sqrt{15}}{2},\frac{1}{2}\right),\left(\frac{\sqrt{15}}{2},\frac{1}{2}\right)$the value of the function is$\frac{17}{4}.$

Thus, the maximum value of the function is $\frac{17}{4}\mathrm{at}\left(−\frac{\sqrt{15}}{2},\frac{1}{2}\right),\left(\frac{\sqrt{15}}{2},\frac{1}{2}\right)$and the minimum value of the function is$-2\mathrm{at}\left(0,-2\right).$