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\centerline{Mathematics 133 -- Intensive Calculus for Science 1}
\centerline{Solutions for Discussion 6}
\centerline{\it November 29, 2001}
\bigskip
\noindent
A.  Calling the length of the straight sides of the track $x$ and
the sides where the semicircles are attached $y$, we get
$$\eqalign{A &= xy + \pi (y/2)^2\cr
           440 = 2x + 2\pi y\cr}$$
We can solve the second equation for $x$ as a function of $y$:
$$x = 220 - \pi y/2,\leqno(1)$$
and substitute into $A$ to get
$$A(y) = 220y - \pi y^2/2 + \pi y^2/4 = 220 y - \pi y^2/4$$
Now we are ready to apply our optimization procedures.  First 
compute the derivative:
$$A'(y) = 220 - \pi y/2$$
so there is one critical point:
$$\eqalign{0 &= 220 - \pi y/2\cr
   \pi y/2 &= 220\cr
   y &= 440/\pi\cr}$$
By the second derivative test, $A''(y) = -\pi/2 < 0$ for all $y$
so this is a {\it local maximum}.  

The problem asks for the dimensions of the track with maximum area,
so we substitute back into (1) to find 
$$x = 220 - (\pi/2)\cdot(440/\pi) = 0$$
So the maximum area track is actually a {\it circle} (the straight sides have
length zero(!)
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\noindent
B.  Let the two numbers be $x,y$.  We have $x + y = 72$ and we want
to maximize $P = xy$.  Solving for $y$ in terms of $x$, $y = 72 - x$,
so $P(x) = x(72-x) = 72x - x^2$.  Then $P'(x) = 72 - 2x$, so 
there is one critical point:  $72-2x = 0$ shows $x = 36$.  This is 
a maximum by the Second Derivative Test, since $P''(x) = -2 < 0$ for all $x$.
Then $y = 72 - 36 = 36$ too, and 
the maximum product is $36\cdot 36 = 1296$.
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\noindent
C.  If you use $x$ inches of the wire to make the circle and $10-x$
inches to make the square, then the circle has circumference $x$,
and the square has perimeter $10-x$.  This shows that the 
radius of the circle is $r = x/(2\pi)$, and the side of the
square is $(10-x)/4$.  Hence the total area (circle plus square) is
$$A(x) = \pi \cdot (x/(2\pi))^2 + ((10-x)/4)^2 = {x^2\over 4\pi} + {(10-x)^2\over 16}$$
Then 
$$A'(x) = {x\over 2\pi} - {10-x\over 8}$$
There is one critical point:
$$\eqalign{0&={x\over 2\pi} - {10-x\over 8}\cr
{10-x\over 8} &= {x\over 2\pi}\cr
20\pi - 2\pi x &= 8 x\cr
10\pi &= (\pi + 4)x \cr
{10\pi\over \pi + 4} &= x\cr}$$
But the second derivative test shows that this critical point is 
actually a {\it local minimum}:  $A''(x) = {1\over 2\pi} + {1\over 8} > 0$.
To find the global maximum, we need to consider the values of $A(x)$
at the endpoints of the range of possible values for $x$:
$A(0) = 100/16$ and $A(10) = 100/(4\pi)$.  We see that $A(10)$ is 
larger, so the maximum area is obtained if we use all of the wire to 
make the circle(!).
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\noindent
D.  The volume of the can is $30 = \pi r^2 h$.  The surface area (top,
bottom, and side wall) is 
$$SA = 2\pi r^2 + 2\pi r h$$
Solve for $h$ from the volume and substitute:
$$SA(r) = 2\pi r^2 + {60\over r}$$
Then 
$$SA'(r) = 4\pi r - {60\over r^2}$$
There is one critical point $r = {\root 3\of 15/\pi} \doteq 1.68$.
The corresponding value of $h$ is $h = 30/(\pi (1.68)^2) \doteq 3.38$.
This is a minimum by the second derivative test:
$$SA''(r) = 4\pi + {120\over r^3}$$ 
is positive at $r \doteq 1.68$  (positive, in fact, for all $r > 0$.)
\bye

