Respuesta :
Take partial derivatives and set them equal to 0:
[tex]\nabla F(x,y)=\langle2x+4,2y-4\rangle=\mathbf 0[/tex]
We find one critical point within the boundary of the disk at [tex](x,y)=(-2,2)[/tex]. The Hessian matrix for this function is
[tex]\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2&0\\0&2\end{bmatrix}[/tex]
which is positive definite, and incidentally independent of [tex]x[/tex] and [tex]y[/tex], so [tex]F(x,y)[/tex] attains a minimum [tex]F(-2,2)=-8[/tex].
Meanwhile, we can parameterize the boundary by [tex]\mathbf r(t)=\langle9\cos t,9\sin t\rangle[/tex] with [tex]0\le t\le2\pi[/tex], which gives
[tex]F(x,y)=F(x(t),y(t))=f(t)=81\cos^2t+81\sin^2t+36\cos t-36\sin t=81+36(\cos t-\sin t)[/tex]
with critical points at
[tex]f'(t)=0\implies 36(-\sin t-\cos t)=0\implies\sin t+\cos t=0\implies t=\dfrac{3\pi}4,\dfrac{7\pi}4[/tex]
At these points, we get
[tex]f\left(\dfrac{3\pi}4\right)=81-36\sqrt2\approx30.0883[/tex]
[tex]f\left(\dfrac{7\pi}4\right)=81+36\sqrt2\approx131.9117[/tex]
so we attain a maximum only when [tex]t=\dfrac{7\pi}4[/tex], which translates to [tex](x,y)=\left(\dfrac9{\sqrt2},-\dfrac9{\sqrt2}\right)[/tex].
[tex]\nabla F(x,y)=\langle2x+4,2y-4\rangle=\mathbf 0[/tex]
We find one critical point within the boundary of the disk at [tex](x,y)=(-2,2)[/tex]. The Hessian matrix for this function is
[tex]\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2&0\\0&2\end{bmatrix}[/tex]
which is positive definite, and incidentally independent of [tex]x[/tex] and [tex]y[/tex], so [tex]F(x,y)[/tex] attains a minimum [tex]F(-2,2)=-8[/tex].
Meanwhile, we can parameterize the boundary by [tex]\mathbf r(t)=\langle9\cos t,9\sin t\rangle[/tex] with [tex]0\le t\le2\pi[/tex], which gives
[tex]F(x,y)=F(x(t),y(t))=f(t)=81\cos^2t+81\sin^2t+36\cos t-36\sin t=81+36(\cos t-\sin t)[/tex]
with critical points at
[tex]f'(t)=0\implies 36(-\sin t-\cos t)=0\implies\sin t+\cos t=0\implies t=\dfrac{3\pi}4,\dfrac{7\pi}4[/tex]
At these points, we get
[tex]f\left(\dfrac{3\pi}4\right)=81-36\sqrt2\approx30.0883[/tex]
[tex]f\left(\dfrac{7\pi}4\right)=81+36\sqrt2\approx131.9117[/tex]
so we attain a maximum only when [tex]t=\dfrac{7\pi}4[/tex], which translates to [tex](x,y)=\left(\dfrac9{\sqrt2},-\dfrac9{\sqrt2}\right)[/tex].
The extreme values of a function are the minimum and the maximum values of the function.
The extreme values are: -8 and 131.91
The given parameters are:
[tex]\mathbf{F(x,y) = x^2 + y^2 + 4x - 4y}[/tex]
[tex]\mathbf{x^2 + y^2 \le 81}[/tex]
Find the gradient of F(x,y)
[tex]\mathbf{f_x(x) = 2x + 4}[/tex]
[tex]\mathbf{f_y(y) = 2y - 4}[/tex]
Set to 0, to solve for x and y
[tex]\mathbf{2x + 4 = 0}[/tex]
[tex]\mathbf{2x = -4}[/tex]
[tex]\mathbf{x = -2}[/tex]
[tex]\mathbf{2y - 4 = 0}[/tex]
[tex]\mathbf{2y = 4}[/tex]
[tex]\mathbf{y = 2}[/tex]
So, the critical point is:
[tex]\mathbf{(x,y) = (-2,2)}[/tex]
Also, we have: [tex]\mathbf{x^2 + y^2 \le 81}[/tex]
Calculate the gradients
[tex]\mathbf{g_x = 2x}[/tex]
[tex]\mathbf{g_y = 2y}[/tex]
Equate the gradients as follows:
[tex]\mathbf{f_x = \lambda \cdot g_x}[/tex]
[tex]\mathbf{f_y = \lambda \cdot g_y}[/tex]
So, we have:
[tex]\mathbf{2x + 4 = \lambda \cdot 2x}[/tex]
[tex]\mathbf{2y - 4 = \lambda \cdot 2y}[/tex]
Make [tex]\mathbf{\lambda}[/tex] the subject, in the above equations
[tex]\mathbf{\lambda = \frac{x + 2}{x}}[/tex]
[tex]\mathbf{\lambda = \frac{y - 2}{y}}[/tex]
Equate the above equations, so we have:
[tex]\mathbf{\frac{x + 2}{x} = \frac{y - 2}{y} }[/tex]
Cross multiply
[tex]\mathbf{xy + 2y = xy - 2x}[/tex]
Subtract xy from both sides
[tex]\mathbf{2y =- 2x}[/tex]
Divide both sides by 2
[tex]\mathbf{y =- x}[/tex]
Substitute [tex]\mathbf{y =- x}[/tex] in [tex]\mathbf{x^2 + y^2 = 81}[/tex]
[tex]\mathbf{x^2 + (-x)^2 = 81}[/tex]
[tex]\mathbf{x^2 + x^2 = 81}[/tex]
[tex]\mathbf{2x^2 = 81}[/tex]
Divide both sides by 2
[tex]\mathbf{x^2 = \frac{81}{2}}[/tex]
Take square roots
[tex]\mathbf{x = \±\frac{9}{\sqrt2}}[/tex]
Rationalize
[tex]\mathbf{x = \±\frac{9\sqrt2}{2}}[/tex]
Recall that: [tex]\mathbf{y =- x}[/tex]
So, we have:
[tex]\mathbf{y = \±\frac{9\sqrt2}{2}}[/tex]
The ordered pairs are:
[tex]\mathbf{(x,y) = \{(\frac{9\sqrt2}{2},-\frac{9\sqrt2}{2}),(-\frac{9\sqrt2}{2},\frac{9\sqrt2}{2})\}}[/tex]
Hence, the points are:
[tex]\mathbf{(x,y) = \{(-2,2),(\frac{9\sqrt2}{2},-\frac{9\sqrt2}{2}),(-\frac{9\sqrt2}{2},\frac{9\sqrt2}{2})\}}[/tex]
Substitute these values in [tex]\mathbf{F(x,y) = x^2 + y^2 + 4x - 4y}[/tex]
[tex]\mathbf{F(2,2) = (-2)^2 + 2^2 + 4(-2) - 4(2) =-8}[/tex]
[tex]\mathbf{F(\frac{9\sqrt2}{2},-\frac{9\sqrt2}{2}) = (\frac{9\sqrt2}{2})^2 + (-\frac{9\sqrt2}{2})^2 + 4(\frac{9\sqrt2}{2}) - 4(-\frac{9\sqrt2}{2}) =131.91}[/tex]
[tex]\mathbf{F(-\frac{9\sqrt2}{2},\frac{9\sqrt2}{2}) = (-\frac{9\sqrt2}{2})^2 + (\frac{9\sqrt2}{2})^2 + 4(-\frac{9\sqrt2}{2}) - 4(\frac{9\sqrt2}{2}) =30.09}[/tex]
Hence, the extreme values of the function are: -8 and 131.91
Read more about extreme values at:
https://brainly.com/question/1286349
