Verified Answer :

LammettHash

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 \(x^2+y^2\le16\).

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

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.

MrRoyal

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

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