91Ó°ÊÓ

The given function is$f(x,y)=3x^2-5y^2$and the constraint is$x^2+y^2≤1.$

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

Let's write the gradient of the given functions,

$$\begin{gathered} \mathrm{∇}f(x,y)=6x\mathbf{i}−10y\mathbf{j} \\ \mathrm{∇}\mathrm{g}(\mathrm{x},\mathrm{y},\mathrm{z})=2\mathrm{x}\mathbf{i}+2\mathrm{y}\mathbf{j} \end{gathered}$$

So, the system of equations is

$$\begin{gathered} \mathrm{∇}f(x,y)=\mathrm{λ}\mathrm{∇}g(x,y) \\ 6x\mathbf{i}−10y\mathbf{j}=\mathrm{λ}(2x\mathbf{i}+2y\mathbf{j}) \end{gathered}$$

Here,$6x=2\mathrm{λ}x,\text{and}−10y=2\mathrm{λ}y.$

The values of $\mathrm{λ},$we get are$\mathrm{λ}=\frac{6x}{2x}=\frac{−10y}{2y}.$

So,

role="math" localid="1649942793601" $$\begin{gathered} 12xy=−20xy \\ 32xy=0 \\ x=0ory=0 \end{gathered}$$

Substitute the above values in the given constraint,

role="math" localid="1649942834541" $$\begin{gathered} x^2+y^2=1 \\ \mathrm{If}x=0,y=1\mathrm{or}\mathrm{if}\mathrm{y}=0,x=1 \end{gathered}$$

Thus, the point where the extreme value appears is$(0,1),(1,0).$

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

So,

When the point is $\left(0,1\right)$the value of the function is $-5.$

And when the point is $\left(1,0\right)$the value of the function is$3.$

Thus, the maximum value of the function is $3\mathrm{at}\left(1,0\right)$and the minimum value of the function is$-5\mathrm{at}\left(0,1\right).$