1
151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
MATH 151-Fall 02
November 19
1. Find two numbers whose difference is 65 and whose product is a minimum.
Let x be the first number and let y be the second number. We want to minimize P = xy. To
do this, we first need the formula in only one variable. Since x − y = 65 (we could have used
y − x = 65) we get that y = x + 65.
P = x(x + 65) = x2 + 65x
P 0 = 2x − 65
P 00 = 2
The critical value for the function P is x = 32.5. We know that there is a local minimum for
the function at this critical value by the second derivative test. Notice the second derivative is
always positive. That means the function will be concave up for any critical value. Hence, a
local min.
The first number is 32.5 and the second number is 97.5.
2. A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other
is bent into a circle. How should the wire be cut so that the total area enclosed is a maximum?
Let x be the length that is cut for the wire to form the square. This means that the rest of the
wire 10 − x is the circumference of the circle.
Now the area enclosed by the wire is A = s2 + πr 2 . Since x is the amount of wire cut for the
x
square, then each side has length of . The circumference of the circle is 2πr = 10 − x. Solving
4
10 − x
for r gives r =
2π
2
x
10 − x 2
A=
+π
16
2π
x2 (10 − x)2
A=
+
16
4π
A0 =
x −(10 − x)
+
8
2π
80
≈ 5.60099
2π + 8
It is very tempting to stop here and say that we are done. However, if you look at a sign chart
for the first derivative or use the second derivative test this critical value will produce a local
min.
1
1
A00 = +
> 0 for all values of x.
8 2π
This critical value can not be a solution. However, thinking about the problem, reveals that
our function A is only defined on the interval [0, 10]. These are the only realistic values that x
can be. Since be have a continuous function on a closed and bounded interval, it will have an
absolute maximum and an absolute minimum at critical values or at the ends of the interval. So
our only choices for our answer are x = 0 or x = 10. If x = 0, then the whole wire is used for
100 2
the circle and will produce a figure that has an area of
m ≈ 7.9577m2 . If x = 10, then the
4π
100 2
whole wire is used for the square and will produce a figure that has an area of
m ≈ 6.25m2 .
16
Answer: Don’t cut the wire and use it to form a circle.
The critical value will be x =
2
151 WebCalc Fall 2002-copyright Joe Kahlig
3. A window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter
of the window is 12ft. find the dimensions of the rectangle that will produce the largest area for
the window.
The variables are shown below on the figure to the left. The perimeter of the window is P =
12 = 3x + 2y. Solving for y gives y = 6 − 1.5x. The area of the window is,
√
√
√
1
x2 3
x 3
x2 3
2
A = xy + ∗ x ∗
= x(6 − 1.5x) +
= 6x − 1.5x +
2
2
4
4
√
√
3
x 3
A0 = 6 − 3x +
and A00 = −3 +
< 0 for all x. Since the second derivative is always
2
2
negative, any critical value will be a local max.
6
√ ≈ 2.811ft.
The critical value is x =
3 − .5 3
Answer: base 2.811ft and the side 1.7835 ft.
√
.5x 3
x
6−x
.5x
y
2
√
4 + x2
x
4. A person in a rowboat 2 miles from the nearest point, called P, on a straight shoreline wishes
to reach a house 6 miles farther down the shore. If the person can row at a rate of 3 miles per
hour and walk at a rate of 5 miles per hour, how far along the shore should the person walk in
order to minimize the amount of time it takes to get to the house?
The variables are shown above in the figure to the right. The time it takes the person to travel
distance
a distance is given by t =
.
rate
√
4 + x2 6 − x
x
1
T =
− The second derivative can be messy to compute,
+
and T 0 = √
2
3
5
5
3 4+x
so we will use the first derivative test (sign chart) to see if the critical value is a minimum.
Now solve for the critical value.
x
1
Now we need to test this critical value.
0= √
−
2
5
3 4 +√
x
Compute T 0 (1) and T 0 (2).
5x − 3 4 + x2
T 0 (1) = −.0509 < 0 implies that for x < 1.5
√
0=
2
15 p
4+x
the function is decreasing.
0 = 5xp
− 3 4 + x2
T 0 (2) = .0357 > 0 implies that for x > 1.5
5x = 3 4 + x2
the function is increasing.
25x2 = 9(4 + x2 )
Thus at x = 1.5 is a local min.
16x2 = 36
3
x=±
Answer: The person should walk 6-1.5 miles
2
or 4.5 miles along the shore.
The only critical value that makes sense, in
the context of this problem, is x = 1.5.