Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software,

Question

19 days ago

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Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software,

graph the function with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x, y) = 9 - 2x + 4y - x2 - 4y2

Mathematics

2 answers:

Zina [9.1K] · Answer 1 · 2026-03-09T02:03:56+03:00

Response:

The point (-1,1/2) serves as a local maximum, with no presence of local minimums or saddle points.

Detailed breakdown:

The function under analysis is $f(x,y) = 9-2x+4y-x^2-4y^2$ , which is a differentiable function concerning variables $x$ and $y$ . To discover potential local maxima, minima, and saddle points, we examine where the partial derivatives equal zero simultaneously. Thus, we need to calculate the partial derivatives:

$\frac{\partial f}{\partial x} = -2-2x = -2(x+1)$

$\frac{\partial f}{\partial y} = 4-8y = 4(1-2y)$

It’s evident that the only critical point identified is (-1,1/2). Notably, the system of equations

$\begin{cases} -2(x+1) &= 0\\ 4(1-2y) &= 0\end{cases}$

yields solutions x= -1 and y=1/2.

To determine if (-1,1/2) is a maximum, minimum, or saddle point, we must calculate the second-order partial derivatives:

$\frac{\partial^2 f}{\partial x^2} = -2 = A$

$\frac{\partial^2 f}{\partial y^2} = -8 = B$

$\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} = 0 = C$ .

Substituting the values into the formula $AB-C^2$ , we compute (-2)(-8)-0=16>0. Since the result is greater than zero and A=-2<0, we conclude that (-1,1/2) is a local minimum.