Group work 2/20 Solutions
1. The frame of a kite is to be made from 6
pieces of wood. The four exterior pieces have
been cut with lengths indicated in the figure.
To maximize the area of the kite, how long
should the pieces of wood be? You do not
need to find the lengths of the board, just set
up the problem until you find the function
you wish to optimize.
Solution:
First, label the edge lengths as follows:
The quantity we wish to maximize is the
area, given by A = xz + yz, as this is the sum
of the area of four triangles.
The quantities a and b are constants, so by
Pythagorean Theorem, we have
x 2 + z 2 = a2
and
y 2 + z 2 = b2 .
In the end, the lengths of the two boards are
x + y and 2z, so we first solve for x, y and z.
Since all quantities are positive, we solve for x and y in terms of z as follows:
√
√
x = a2 − z 2
and
y = b2 − z 2 .
Since the area A is what we are trying to optimize, we substitute to write A in terms of
the single variable z:
√
√
A(z) = a2 − z 2 (z) + b2 − z 2 (z).
We wish to find values of z such that A0 (z) = 0.
Using the product rule, we have
√
√
z2
z2
A0 (z) = a2 − z 2 − √
+ b2 − z 2 − √
.
a2 − z 2
b2 − z 2
This becomes
a2 − z 2 − z 2 b 2 − z 2 − z 2
A0 (z) = √
+ √
.
a2 − z 2
b2 − z 2
We now must solve
a2 − 2z 2
b2 − 2z 2
√
+√
= 0.
a2 − z 2
b2 − z 2
1
2. A man launches his boat from a point A on a bank of a straight river, 5km wide, and
wants to reach point B, 5km downstream on the opposite bank as quickly as possible. He
could row his boat directly across the river to point C and then run to B, or he could row
directly to B, or he could row to some point D between C and B and then run to B. If he
can row 6km/h and run 10km/h where should he land to reach B as soon as possible?
Hint: At the end if you are not sure how to compare fractions, try their squares.
Solution:
Step 1: Interpret the problem
If we let x be the distance from C to D, then the running distance from D to B is√5 − x and
the Pythagorean Theorem gives the rowing distance from point A to point D as x2 + 25.
Step 2: Find an Equation
√
The time it will take him to row across the lake is x2 + 25/6 and the running time is
(5 − x)/10 so the total time as a function of x is
√
x2 + 25 5 − x
T (x) =
+
6
10
Step 3: Feasible Domain
2
The domain of this function T is [0, 5].
Step 4: Maximize your equation
We find the critical points through the derivative
−1
x
+
T 0 (x) = √
2
10
6 x + 25
Setting this equal to 0 we find that
x
−1
√
+
= 0
2
10
6 x + 25
√
5x = 3 x2 + 25
T (0) =
4
3
25x2 = 9(x2 + 25)
⇒
16x2 = 9(25)
The only critical number is x =
x
1
√
=
2
10
6 x + 25
⇒
⇒
x =
15
4
15
4
so using the closed interval method we evaluate
√
15
5 2
7
T
T (5) =
=
4
6
6
To compare these we see that
49
50
T (5) =
>
= T
36
36
2
So the minimum occurs at
T
15
4
=
3
7
.
6
15
4
2