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Course Notes for Multivariable Calculus

Remkes Kooistra

Section 5.2 Week 5 Activity

Subsection 5.2.1 Extrema

Activity 5.2.1.

Calculate and classify the extrema of the function
\begin{equation*} f(x,y) = \sqrt{xy + 1} \end{equation*}
Solution.
First I calculate the gradient.
\begin{align*} \nabla f(x,y) \amp = \left( \frac{y}{2\sqrt{xy+1}}, \frac{x}{2\sqrt{xy+1}} \right) \end{align*}
Then I set the gradient equal to zero to find the critical points. This involves solving a system of two equations.
\begin{align*} \frac{y}{2\sqrt{xy+1}} \amp = 0 \\ \frac{x}{2\sqrt{xy+1}} \amp = 0 \end{align*}
This is an easy system to solve. The fractions are zero only when their numerators are zero, so the only possibilities here are \(x=0\) and \(y=0\text{.}\) This gives the critical point \((0,0)\text{.}\) Then I calculate the determinant of the Hessian and evaluate it at the critical point.
\begin{align*} D \amp = \frac{\del^2 f}{\del x^2} \frac{\del^2 f}{\del y^2} - \left( \frac{\del^2 f}{\del x \del y} \right)^2 \\ D \amp = \left( \frac{-y^2}{4(xy+1)^{\frac{3}{2}}} \right) \left( \frac{-x^2}{4(xy+1)^{\frac{3}{2}}} \right) - \left( \frac{xy+2}{4(xy+1)^{\frac{3}{2}}} \right)^2 \\ D(0,0) \amp = 0 - \left( \frac{2}{4} \right)^2 = \frac{1}{4} \end{align*}
The Hessian determinant is negative, so the critical point at \((0,0)\) is a saddle point.

Activity 5.2.2.

Calculate and classify the extrema of the function
\begin{equation*} f(x,y) = (x^2 - 1) \ln (xy) \end{equation*}
Solution.
First I calculate the gradient.
\begin{align*} \nabla f(x,y) \amp = \left( 2x \ln xy + (x^2 - 1) \frac{1}{x}, (x^2-1) \frac{1}{y} \right) \\ \amp = \left( 2x \ln (xy) + x - \frac{1}{x}, \frac{x^2 - 1}{y} \right) \end{align*}
Then I set the gradient equal to zero to find the critical points. This involves solving a system of two equations.
\begin{align*} 2x \ln (xy) + x - \frac{1}{x} \amp = 0 \\ \frac{x^2 - 1}{y} \amp = 0 \end{align*}
I work with the second equation first. The numerator of this fraction needs to be zero, so \(x^2 = 1\text{.}\) This gives two possible \(x\) coordinates: \(x = \pm 1\text{.}\) For each of these \(x\) coordinates, I can replace \(x\) in the first equation and solve for \(y\text{.}\) First I substitute \(x = 1\text{.}\)
\begin{equation*} 2 \ln (y) + 1 - 1 = 0 \implies \ln y = 0 \implies y = 1 \end{equation*}
Then I substitute \(x = -1\)
\begin{equation*} -2 \ln (-y) - 1 + 1 = 0 \implies -\ln (-y) = 0 \implies y = -1 \end{equation*}
This gives me the critical points \((1,1)\) and \((-1,-1)\text{.}\) Then I calculate the determinant of the Hessian and evaluate it at the critical point.
\begin{align*} D \amp = \frac{\del^2 f}{\del x^2} \frac{\del^2 f}{\del y^2} - \left( \frac{\del^2 f}{\del x \del y} \right) \\ D \amp = \left( 2 \ln (xy) + 2 + 1 + \frac{1}{x^2} \right) \left( \frac{1-x^2}{y^2} \right) - \left( \frac{2x}{y} \right)^2 \\ D(1,1) \amp = \left( 0 + 3 + 1 \right) \left( 0 \right) - \left( \frac{2}{1} \right)^2 = -4 \\ D(-1,-1) \amp = \left( 0 + 3 + 1 \right) \left( 0 \right) - \left( \frac{2}{1} \right)^2 = -4 \end{align*}
The Hessian determinant is negative for points points, so they are both saddle points.

Activity 5.2.3.

Calculate and classify the extrema of the function
\begin{equation*} f(x,y) = \sin (x) \cos (y) \end{equation*}
Solution.
First I calculate the gradient.
\begin{align*} \nabla f(x,y) \amp = \left( \cos (x) \cos (y), - \sin (x) \sin (y) \right) \end{align*}
Then I set the gradient equal to zero to find the critical points. This involves solving a system of two equations.
\begin{align*} \cos x \cos y \amp = 0 \\ -\sin x \sin y \amp = 0 \end{align*}
This is a complicated system. To make both terms vanish, I have to use \(x\) and \(y\) as the zeros of sine or cosine in matching pairs. This gives two families of point. The first is
\begin{align*} \amp \left( n\pi, (2m+1) \frac{\pi}{2} \right) \amp \amp n,m \in \ZZ \end{align*}
For this first family, the \(x\) coordinate makes the sine term vanish and the \(y\) coordinate makes the cosine term vanish. The second family does the opposite.
\begin{align*} \amp \left( (2n + 1) \frac{\pi}{2}, m\pi \right) \amp \amp n,m \in \ZZ \end{align*}
Here, the \(x\) coordinate makes the cosine terms vanish and the \(y\) coordinate makes the sine terms vanish. Then I calculate the determinant of the Hessian and evaluate it at the critical points (which I will have to do in families, since there are infinitely many critical points).
\begin{align*} D \amp = \left( -\sin x \cos y \right) \left( -\sin x \cos y \right) - \left( - \cos x \sin y \right)^2 \\ \amp = \sin^2 x \cos^2 y - \cos^2 x \sin^2 y \end{align*}
I have to analyse the two families of critical points. For the points in the family \(\left( n\pi, (2m+1) \frac{\pi}{2} \right)\text{,}\) \(D = -1\text{,}\) so all of these points are saddle points. For points in the family \(\left( (2n+1) \frac{\pi}{2}, m \pi \right)\text{,}\) \(D = 1\text{,}\) so these are minima or maxima. I need to additionally look at \(\frac{\del^2 f}{\del x^2} = -\sin x \cos y\text{.}\) If \(m,n\) are both odd or both even, this partial derivative is negative and the critical point is a maximum. If one of \(m,n\) is odd and the other is even, this partial derivative is positive and the critical point is a minimum.

Activity 5.2.4.

Calculate and classify the extrema of the function
\begin{equation*} f(x,y) = x^3y + 4x^2y + xy - 6y \end{equation*}
Solution.
First I calculate the gradient.
\begin{align*} \nabla f(x,y) \amp = (3x^2y + 8xy + y, x^3 + 4x^2 + x - 6) \end{align*}
Then I set the gradient equal to zero to find the critical points. This involves solving a system of two equations.
\begin{align*} 3x^2 y + 8xy + y \amp = 0 \\ x^3 + 4x^2 + x - 6 \amp = 0 \end{align*}
I can work with the second equation first, since it only involves the \(x\) coordinate. This cubic has three roots \(x = 1, -2, -3\text{.}\) Then I look at the second equation, which I can factor.
\begin{equation*} y(3x^2 + 8x + 1) = 0 \end{equation*}
I need to replace \(x\) with the three roots. However, none of the three roots will produce zero in the second factor in this equation. Therefore, to satisfy the first equation, I must set \(y=0\text{.}\) This gives me the critical points \((1,0)\text{,}\) \((-2,0)\) and \((-3,0)\text{.}\) Then I calculate the determinant of the Hessian and evaluate it at the critical points.
\begin{align*} D \amp = \frac{\del^2 f}{\del x^2} \frac{\del^2 f}{\del y^2} - \left( \frac{\del^2 f}{\del x \del y} \right) \\ D \amp = (6xy + 8y)(0) - (3x^2 + 8x + 1)^2 = -(3x^2 + 8x + 1)^2 \end{align*}
\(D\) will always be negative, regardless of where I evaluate it. Therefore, all three points are saddlepoints.

Activity 5.2.5.

Calculate and classify the extrema of the function
\begin{equation*} f(x,y) = x^2y^2 - 3xy^2 -4x^2 \end{equation*}
Solution.
First I calculate the gradient.
\begin{align*} \nabla f(x,y) \amp = (2xy^2 - 3y^2 - 8x, 2x^2y - 6xy) \end{align*}
Then I set the gradient equal to zero to find the critical points. This involves solving a system of two equations.
\begin{align*} 3xy^2 - 3y^2 - 8x \amp = 0 \\ 2x^2 y - 6xy \amp = 0 \end{align*}
I work with the second equation first. I can factor the left side.
\begin{equation*} (x)(y)(2x - 6) = 0 \end{equation*}
This leads to three possibilities: \(x=0\text{,}\) \(y=0\) and \(x=3\text{.}\) I need to try all these possibilities in the first equation. Let me start with \(x=0\) and rewrite the first equation.
\begin{equation*} 3(0)y^2 - 3y^2 - 8(0) = 0 \implies 3y^2 = 0 \end{equation*}
This is solved only when \(y=0\text{,}\) so I get a critical point at \((0,0)\text{.}\) If i start with \(x=0\text{,}\) I get the same critical point. Lastly, I need to check \(x=3\text{.}\)
\begin{equation*} 3(3)y^2 - 3y^2 - 8(3) = 0 \implies 6y^2 = 24 \implies y^2 = 8 \end{equation*}
There are two solutions here: \(y = \pm 2\sqrt{2}\text{.}\) That gives two more critical points. In total, I have three critical points \((0,0)\text{,}\) \((3, 2\sqrt{2})\) and \((3, -2\sqrt{2})\text{.}\) Then I calculate the determinant of the Hessian and evaluate it at the critical point.
\begin{align*} D \amp = \frac{\del^2 f}{\del x^2} \frac{\del^2 f}{\del y^2} - \left( \frac{\del^2 f}{\del x \del y} \right) \\ D \amp = (2y^2 - 8)(2x^2 - 6y) - (4xy - 6y)^2 \\ D(0,0) \amp = 0 \\ D(3,2\sqrt{2}) \amp = -240 \\ D(3,-2\sqrt{2}) \amp = -240 \end{align*}
The test is inconclusive for \((0,0)\text{.}\) By looking at the graph of the function, I could guess that there is a saddle point at the origin, but it is difficult to see this directly from the function. The Hessian determinant is negative for the other two critical points, so these are certainly both saddle points.

Subsection 5.2.2 Conceptual Review Questions

  • What is a saddle point?
  • How does optimization in several variables differ from the single variable situation?
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