Question

Find the extreme values of f on the region described bythe inequality.
f(x,y) =2x2 + 3y2 - 4x -2, x2 + y2 ≤ 16
_______(maximum)
________ (minimum)

Answer 1

Let

$f(x,y)=2x^2+3y^2-4x-2$
$g(x,y)=x^2+y^2-16$

so that the Lagrangian is

$L(x,y,\lambda)=f(x,y)-\lambda g(x,y)=(2-\lambda)x^2+(3-\lambda)y^2-4x-2+16\lambda$

You have

$\begin{cases}L_x=-2x\lambda+4x-4=0\\L_y=6y-2y\lambda=0\\L_\lambda=16-x^2-y^2=0\end{cases}$

Solving for $(x,y)$, you find four potential critical points, each giving values of $f(x,y)$ listed below:

$(-4,0)\implies f(-4,0)=46$
$(-2,-2\sqrt3)\implies f(-2,-2\sqrt3)=50$
$(-2,2\sqrt3)\implies f(-2,2\sqrt3)=50$
$(4,0)\implies f(4,0)=14$

Before you choose any of these, keep in mind that $f(x,y)$ is not a plane, so you need to also check for any possible critical points within the disk $x^2+y^2\le16$.

$\nabla f(x,y)=\left\langle4x-4,6y\right\rangle$

The gradient is zero when $x=1$ and $y=0$. At this point, you have $f(1,0)=-4$.

So, there are two maxima, both 50, and one minimum, -4.

Answer 2

The extreme values of a function are the minimum and the maximum values of the function.

The minimum and the maximum values of f are: -4 and 46, respectively.

The given parameters are:

$\mathbf{f(x,y) = 2x^2 + 3y^2 - 4x - 2}$

$\mathbf{x^2 + y^2 \le 16}$

Find the gradient of F(x,y).

$\mathbf{f_x(x) = 4x - 4}$

$\mathbf{f_y(y) = 6y}$

Set to 0, to solve for x and y

$\mathbf{4x - 4 = 0}$

$\mathbf{4x = 4}$

$\mathbf{x = 1}$

$\mathbf{6y = 0}$

$\mathbf{y = 0}$

So, the critical points are:

$\mathbf{(x,y) = (1,0)}$

Calculate the gradients of $\mathbf{x^2 + y^2 \le 16}$

$\mathbf{g_x = 2x}$

$\mathbf{g_y = 2y}$

Equate the gradients as follows

$\mathbf{f_x(x) = \lambda g_x(x)}$

$\mathbf{f_y(y) = \lambda g_y(y)}$

So, we have:

$\mathbf{4x - 2 = \lambda \cdot 2x}$

$\mathbf{6y = \lambda \cdot 0}$

$\mathbf{6y = 0}$

Divide both sides by 6

$\mathbf{y = 0}$

Substitute $\mathbf{y = 0}$ in $\mathbf{x^2 + y^2 = 16}$

$\mathbf{x^2 + 0^2 = 16}$

$\mathbf{x^2 = 16}$

Take square roots

$\mathbf{x = \pm4}$

So, we have:

$\mathbf{(x,y) = \{(1,0)(4,0)(-4,0)\}}$

Substitute these values in $\mathbf{f(x,y) = 2x^2 + 3y^2 - 4x - 2}$

$\mathbf{f(1,0) = 2(1)^2 + 3(0)^2 - 4(1) - 2 =-4}$

$\mathbf{f(4,0) = 2(4)^2 + 3(0)^2 - 4(4) - 2 =14}$

$\mathbf{f(-4,0) = 2(-4)^2 + 3(0)^2 - 4(-4) - 2 = 46}$

Hence, the minimum and the maximum values of f are: -4 and 46, respectively.

Read more about extreme values at:

brainly.com/question/1286349