Lesson Tricky solution to find the maximum of a function defined by a complicated expression

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<H2>Tricky solution to find the maximum of a function defined by a complicated expression</H2>

<H3>Problem 1</H3>A company has &nbsp;$33600 &nbsp;to spend on the development and promotion of a new product. 
The company estimates that if &nbsp;x &nbsp;is spent on development and &nbsp;y &nbsp;is spent on promotion, 
then approximately &nbsp;{{{((x/2)(y^3/2))/400000}}} &nbsp;items of new product will be sold. 
Based on this estimate, &nbsp;how much should the company spend on development 
so the maximum number of products will be sold?

<B>Solution</B>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It is clear that this problem is on finding maximum.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It can be solved by applying standard Calculus to the given function, 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;but it is, &nbsp;obviously, &nbsp;not very pleasant exersize, &nbsp;taking into account 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;complicated formula for the function.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;There is a trick, &nbsp;and I will show it below, &nbsp;which makes the solution an easy and pleasant 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;procedure and makes a reader happy, &nbsp;since he &nbsp;(or she) &nbsp;learns something new.

<pre>
Let  u = (x^(1/2))*(y^(3/2))/400000 be the given function;

v = (x^(1/2))*(y^(3/2));

w = {{{x*y^3}}}.

It is obvious that "u" and "v" have the maximum at the same point (x,y)
(dividing by the constant value 400,000 does not shift the position of maximum on (x,y)-plane).

Which is even more interesting and productive, is the fact that "v" and "w" also have 
the maximum at the same point (x,y) on the coordinate plane.

+-------------------------------------+
        |  It is because  w(x,y) = v^2(x,y).  |
        +-------------------------------------+

Now our task is much more simple: find the point (x,y) on the coordinate plane such that

x + y = 33600    (1)

and

maximizes  w(x,y) = {{{x*y^3}}},  x >=0,  y >= 0.    (2)

To do it,  from (1)  I express  y = 33600-x  and substitute it into (2).

Then I get the problem to maximize

W(x) = {{{x*(33600-x)^3}}}.

Apply a standard Calculus procedure: find the derivative and equate it to zero.
You will get then

(33600-x)^3 = {{{3x*(33600-x)^2}}},  0 < x < 33600.

Simplify by reducing the factor  {{{(33600-x)^2}}} in both sides

33600-x = 3x

33600 = 3x + x = 4x

x = 33600/4 = 8400.

So, the maximum of w(x,y),  v(x,y)  and  u(x,y)  is  achieved at x= 8400,  y= 33600-8400 = 25200

max u(x,y) = u(8400,25200) = (8400^(1/2)*25200^(3/2)))/400000 = 916.6  (rounded)

<U>ANSWER</U>.  The point of maximum is  (x,y) = (8400,25200).

The value of the maximum is about 916.6.

The amount to spend on development is x = $8400,  leaving  $25200 to spend on promotion.
</pre>

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