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find all points where the given function has any relative extrema. iden…

Question

find all points where the given function has any relative extrema. identify any saddle points.
$f(x,y)=-5x^{2}-6xy - 5y^{2}-44x - 20y + 8$
select the correct choice below and fill in any answer boxes within your choice.
a. the relative maximum/maxima is/are located at
(simplify your answers. type ordered pairs. use a comma to separate answers as needed.)
b. there are no relative maxima.

Explanation:

Step1: Find first - order partial derivatives

The first - order partial derivatives of \(z = f(x,y)=-5x^{2}-6xy - 5y^{2}-44x-20y + 8\) are:
\(f_{x}=\frac{\partial f}{\partial x}=-10x-6y - 44\)
\(f_{y}=\frac{\partial f}{\partial y}=-6x-10y - 20\)

Step2: Solve the system of equations \(f_{x}=0\) and \(f_{y}=0\)

We have the system of equations:
\(

$$\begin{cases}-10x-6y-44 = 0\\-6x - 10y-20=0\end{cases}$$

\)
From the first equation \( - 10x-6y=44\), we get \(y=\frac{-10x - 44}{6}\)
Substitute \(y=\frac{-10x - 44}{6}\) into the second equation:
\(-6x-10\times\frac{-10x - 44}{6}-20 = 0\)
Multiply through by \(6\) to clear the fraction:
\(-36x+100x + 440-120 = 0\)
\(64x=-320\), so \(x=- 5\)
Substitute \(x = - 5\) into \(y=\frac{-10x - 44}{6}\), \(y=\frac{-10\times(-5)-44}{6}=\frac{50 - 44}{6}=1\)

Step3: Find second - order partial derivatives

\(f_{xx}=\frac{\partial^{2}f}{\partial x^{2}}=-10\), \(f_{xy}=\frac{\partial^{2}f}{\partial x\partial y}=-6\), \(f_{yy}=\frac{\partial^{2}f}{\partial y^{2}}=-10\)

Step4: Use the second - derivative test

The discriminant \(D=f_{xx}f_{yy}-(f_{xy})^{2}\)
\(D=(-10)\times(-10)-(-6)^{2}=100 - 36 = 64\)
Since \(D>0\) and \(f_{xx}=-10<0\)

Answer:

A. The relative maximum/maxima is/are located at \((-5,1)\)