Question

Consider the function
f(x,y) = y\sqrt x - y^2 - 3 x + 11 y.
Find and classify all critical points of the function. If there are more blanks than critical points, leave the remaining entries blank.
f_x =
f_y = f_{xx} = f_{xy} = f_{yy} = The critical point with the smallestx-coordinate is
(__,__) Classification:
(local minimum, local maximum, saddle point, cannot be determined)
The critical point with the next smallestx-coordinate is
(__,__) Classification:
(local minimum, local maximum, saddle point, cannot be determined)
The critical point with the next smallestx-coordinate is
(__,__) Classification:
(local minimum, local maximum, saddle point, cannot be determined)

Answer 1

By using the concept of maxima,  it can be  determined that

$f_x(x, y) = \frac{y}{2\sqrt{x}} - 3$

$f_y(x, y) = \sqrt{x} - 2y +11$

$f_{xx}(x, y) = -\frac{1}{4}x^{-\frac{3}{2}}$

$f_{yy} = -2$

$f_{xy} = \frac{1}{2\sqrt{x}}$

Critical point with the smallest x - coordinate

(0, $\frac{11}{2}$)

Classification cannot be determined

Critical point with the next smallest x - coordinate

(1, 6)

Classification is local maxima

What is maxima of a function?

Maxima of a function gives the maximum value of a function in a given interval or within the whole domain.

f(x, y) = $y\sqrt{x} -y^2 -3x + 11y$

$f_x(x, y) = \frac{y}{2\sqrt{x}} - 3$

$f_y(x, y) = \sqrt{x} - 2y +11$

Putting x = 0 in $f_y(x, y) = 0$,

-2y + 11 = 0

$y = \frac{11}{2}$

$(0, \frac{11}{2})$ is a critical point

Putting $f_x(x, y) = 0$ and $f_y(x, y) = 0$

$\frac{y}{2\sqrt{x}} - 3 = 0\\\frac{y}{2\sqrt{x}} = 3\\y = 6\sqrt{x}$

$\sqrt{x} - 2y+11 = 0\\\sqrt{x} -2\times 6\sqrt{x}+11=0\\\sqrt{x}-12\sqrt{x} + 11 = 0\\11\sqrt{x} = 11\\\sqrt{x} = \frac{11}{11}\\\sqrt{x} = 1\\x = 1$

When x = 1, y = $6 \times \sqrt{1} = 6$

(1, 6) is a critical point

$f_{xx}(x, y) = -\frac{1}{4}x^{-\frac{3}{2}}$

$f_{yy} = -2$

$f_{xy} = \frac{1}{2\sqrt{x}}$

The point (0, $\frac{11}{2}$) is undefined in the second order partial derivative

For the critical point (1, 6)

$f_{xx}(1, 6) = -\frac{1}{4}\\f_{yy}(1, 6) = -2\\f_{xy}(1, 6) = \frac{1}{2}$

$f_{xx}f_{yy}-(f_{xy})^2$

$-\frac{1}{4}\times -2-(\frac{1}{2})^2\\\frac{1}{2} - \frac{1}{4}\\\frac{1}{4} > 0$

$f_{yy} = -2 < 0$

Hence (1, 6) is a point of local maxima

So it can be  determined that

$f_x(x, y) = \frac{y}{2\sqrt{x}} - 3$

$f_y(x, y) = \sqrt{x} - 2y +11$

$f_{xx}(x, y) = -\frac{1}{4}x^{-\frac{3}{2}}$

$f_{yy} = -2$

$f_{xy} = \frac{1}{2\sqrt{x}}$

Critical point with the smallest x - coordinate

(0, $\frac{11}{2}$)

Classification cannot be determined

Critical point with the next smallest x - coordinate

(1, 6)

Classification is local maxima

To learn more about maxima, refer to the link-

https://brainly.com/question/82347

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