WORKSHOP 2·12
Solutions
Warm-up
Let f (x) be a positive function with one critical point (which happens to be a local
minimum) at x = a. Now consider the function g(x) = f (x)2 . Explain why g(x) has the
same critical point and minimum as f (x).
Solution:
Since f (x) has a local minimum at x = a, then f 0 (a) = 0 and f 00 (a) > 0. f (x) is also a positive function, so
f (a) > 0. Now let’s find the firt and second derivative of g(x), using the chain rule.
d
(f (x))2 = 2f (x)f 0 (x)
dx
d
(2f (x)f 0 (x)) = 2f (x)f 00 (x) + 2(f 0 (x))2 .
g 00 (x) =
dx
g 0 (x) =
Checking at x = a we obtain
g 0 (a) = 2f (a)f 0 (a) = 0
g 00 (a) = 2f (a)f 00 (a) > 0,
thus we have a critical point at a and it is a minimum.
Main Problem
You need to fire a rocket from somewhere along the ground, past the obstruction, to hit
the target. You may assume that the rocket travels in a straight line. Find x such that
your trajectory length L is as short as possible (hence your rocket needs as little fuel as
possible). Once you have done so, cut out a piece of masking tape with the exact length
of this predicted shortest path.
Obstacle (h)
Target (A)
L
d
Solution:
We wish to find the absolute minimum of L. Of all the quantities in the problem the one we directly control
is the distance x, so we wish to express L as a function of x and constants only.
Let A be the height the rocket hits the target, then we may use the Pythagorean theorem to find L
L2 = A2 + (d + x)2 .
(1)
From the warm-up we know that the local minimums of L and L2 are the same, so we will continue working
with L2 from now on.
In the previous equation (1) the quantity A also depends on x. The closer we are to the target the higher
the rocket is going to hit the target. We may use similar triangles to find an expression for A.
h
A
=
d+x
x
h(d + x)
A=
.
x
Substituting into (1) we get
2
h(d + x)
L =
+ (d + x)2
x
h2
2
L = 1 + 2 (d + x)2 .
x
2
To find its critical points we find the first derivative and set it equal to zero
−2h2
h2
d 2
2
L =
(d + x) + 2 1 + 2 (d + x) = 0
dx
x3
x
2
2
h
h
1 + 2 (d + x) = 3 (d + x)2 (∗)
x
x
2
h
h2
1 + 2 = 3 (d + x)
x
x
x3 + h2 x = h2 (d + x)
X 3 = h2 d
1
x = (h2 d) 3 .
From the second line (∗) we also have x = −d as a critical point, but it won’t be inside our domain. The
domain of our function is 0 < x < ∞. From observing our domain, if we approach x = 0 the rocket will hit
the target higher and higher and the limit will be positive infinity. similarly if x approaches positive infinity
1
so will L2 and L. Thus, we expect our critical point x = (h2 d) 3 to be the global minimum of L. To verify
this we compute the second derivative
d2 2
6h2 (d + x)2
−4h2 (d + x) −4h2 (d + x)
2h2
L =
−
−
+2+ 2
2
4
3
3
dx
x
x
x
x
6h2 (d + x)2 − 8h2 (d + x)x + 2h2 x2 + 2x4
=
x4
2x4 + 4h2 xd + 6h2 d2
=
> 0.
x4
Thus the function is concave up, we have a local minimum that is also a global minimum.