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\centerline{Mathematics 41, section 1 -- Multivariable Calculus}
\centerline{Solutions for Practice Exam 3}
\bigskip
\noindent
I.
\item{A)}  For $f(x,y) = x^3 + y^2 - 3x$, we have
$${\partial f \over \partial x} = 3x^2 - 3\qquad
{\partial f \over \partial y} = 2y$$
The critical points are where both partial derivatives equal
zero:
$$(x,y) = (\pm 1, 0)$$
From the gradient vector field:
\vglue 2.0 truein
we see that $(1,0)$ is a {\it source} for $\nabla f$
and $(-1,0)$ is a {\it saddle point}.  By the First Derivative Test
(see page 293 of text), this means $(1,0)$ is a local minimum
and $(-1,0)$ is a saddle point.
\item{B)}  The second order partial derivatives are
$${\partial^2 f \over \partial x^2} = 6x \qquad
{\partial^2 f\over \partial x \partial y} = 0\qquad
{\partial^2 f \over \partial y^2} = 2$$
We have 
$$H = {\partial^2 f \over \partial x^2}{\partial^2 f \over \partial y^2} - 
\left({\partial^2 f\over \partial x \partial y}\right)^2 = 12x - 0^2 = 12x$$
So at $(1,0)$, $H = 12 > 0$ and ${\partial^2 f \over \partial x^2} = 6 > 0$
which shows $(1,0)$ is a local minimum.  At $(-1,0)$, $H = -12 < 0$
so this is a saddle point.  (This also agrees with A!)
\item{C)}  The Lagrange Multiplier equations are:
$$\eqalign{
3x^2 - 3 &= 2\lambda x\cr
2y &= 2\lambda y\cr
x^2 + y^2 &= 9\cr}$$
From the second equation, we get $2y(1 - \lambda) = 0$, so 
either $y = 0$, or $\lambda = 1$.  
\medskip
\item{} If $y = 0$, then from the constraint equation, $x = \pm 3$,
so we get two points $(\pm 3, 0)$.  
\medskip
\item{}If $\lambda = 1$, then from the first equation, 
$3x^2 - 3 = 2x$, or $3x^2 - 2x - 3 = 0$.  Using the quadratic
formula we get
$$x = {2 \pm \sqrt{4 - 4\cdot 3\cdot 3}\over 6} = {1 \pm \sqrt{10}\over 3}
\doteq -.72, 1.39$$
With $x = -.72$, we get from the constraint equation $y \doteq \pm 2.91$;
with $x = 1.39$, we get $y \doteq \pm 2.66$.
Comparing values of $f$, we get
$$\matrix{
&f(3,0) &= 18 &f(-3,0) &= -18\cr
&f(-.72,\pm 2.91) &\doteq 10.25 &&\cr
&f(1.39,\pm 2.66) &\doteq 5.59 &&\cr}$$
The maximum and minimum values are the 18, -18 respectively.
\item{D)}  The maximum and minimum values are the 18, -18 respectively
(compare with $f(1,0) = -2$ found in part A).
\bigskip
\noindent 
II.  
\item{A)} Subdividing the rectangle into 9 equal smaller rectangles
(so that we can use all the data points), we have $\Delta x = \Delta y = 2/3$.
The Riemann sum is found by adding up all the table entries
times the area of the small rectangle containing the given point:
$$6 \cdot {2\over 3} \cdot {2\over 3} + 8 \cdot {2\over 3} \cdot {2\over 3}
+\cdots + 6 \cdot {2\over 3} \cdot {2\over 3} = 68/3$$
\item{B)}  
$$\eqalign{
\int_1^3\int_0^2 2x + y^2\ dy\ dx &= \int_1^3 (2xy + y^3/3)|_0^2\ dx\cr
                                  &= \int_1^3 4x + 8/3\ dx\cr
                                  &= 2x^2 + 8x/3|_1^3\cr
                                  &= 64/3}$$
\bigskip
\noindent
III.
\item{A)}  We must split the region at $x = 0$, since the 
lower edge is $y = -x$ to the left of $x = 0$ and the lower edge
is $y = x$ to the right.
$$\int_{-3/\sqrt{2}}^0 \int_{-x}^{\sqrt{9-x^2}} y\ dy\ dx +
\int_0^{3/\sqrt{2}} \int_{x}^{\sqrt{9-x^2}} y\ dy\ dx$$
\item{B)}  We must also split the region here since the 
left and right boundaries are the lines $x = -y$ and $x = y$ below
$y = 3\/sqrt{2}$ and the left and right halves of the circle above:
$$\int_0^{3/\sqrt{2}} \int_{-y}^y y\ dx\ dy + 
\int_{3/\sqrt{2}}^3 \int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} y\ dx\ dy$$
\item{C)}  Using either form, the value is
$9\sqrt{2}$.
\bye


