Maximizing Horizontal Distance

Date: 04/10/98 at 12:44:47
From: Graham Carter
Subject: Algebra/Trigonometry

I've been trying to prove that without air resistance, an object needs 
to be thrown at 45 degrees to obtain the furthest possible horizontal 
distance. Effectively, I can prove this by stating that the product of 
sin(A) and cos(A) is greatest when they are the same; ie: 1/sqrt(2). 

However, I'm finding it difficult to prove this with algebra. Also 
related to this, I would like to know how to show that if the sum of 
two numbers is always constant, the product is greatest when they are 
the same. If you could help, that would be great.

Thanks, Graham Carter

Date: 04/10/98 at 14:35:09
From: Doctor Rob
Subject: Re: Algebra/Trigonometry

To answer your second question first, you want to maximize u*v, 
subject to the constraint that u + v = c, a constant. That is the same 
as maximizing u*(c-u) = c^2/4 - (c/2-u)^2 (completing the square!), 
which is the same as minimizing (c/2-u)^2, and the minimum of this is 
clearly zero, which happens when u = c/2 = v.

The path of the projectile is given by the equations

   y = -g*t^2/2 + k*sin(A)*t
   x = k*cos(A)*t

Here g is the acceleration of gravity, and k is the object's initial 
speed. Then y = 0 when t = 0 and when t = 2*k*sin(A)/g, and this is 
the time of impact, at which time the x-value, that is the range, is:

   x = 2*k^2*sin(A)*cos(A)/g

To maximize this, you need to maximize sin(A)*sqrt[1-sin^2(A)], which 
is the same as maximizing its square:

   sin^2(A)*[1-sin^2(A)] = 1/4 - [1/2 - sin^2(A)]^2

This is maximized when [1/2 - sin^2(A)]^2 is minimized. The minimum 
value of any square is zero. In this case, it means that
sin(A) = 1/sqrt(2), so cos(A) = 1/sqrt(2), and the maximum range is 
2*k^2*sin(A)*cos(A)/g = k^2/g. Note sin(A) = cos(A) = 1/sqrt(2) means 
the angle of fire is 45 degrees.

This is all done without calculus, which would have helped.

A simpler method is to use a trigonometric identity for sin(2*A) to 
rewrite the range equation as:

   x = k^2*sin(2*A)/g

and this is maximized when sin(2*A) = 1, so 2*A = 90 degrees and
A = 45 degrees, and the maximum range is k^2/g.

-Doctor Rob,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/