Question 243614
Find two positive numbers such the sum of the first and twice the second is 100 and their product is as large as possible?
<pre><font size = 4 color = "indigo"><b>
Let t = the first number
Let x = the second number

>>...the sum of the first and twice the second is 100...<<

{{{t+2x = 100}}}

We want to maximize the product xy.  

Let y = the product = tx

{{{y=tx}}}

Solve {{{t+2x = 100}}} for t
      {{{t= 100-2x}}}

Substitute {{{100-2x}}} for t in

{{{y = tx}}}

{{{y = (100-2x)x}}}

{{{y = x(100-2x)}}}

{{{y = 100x-2x^2}}}

If you are taking algebra, not calculus, then find the
vertex of the parabola

{{{y=-2x^2+100x+0}}}

Since the coefficient of {{{x^2}}} is negative, the
graph opens downward and thus reaches a maximum at the
vertex.

The formula for the x-coordinate of the vertex = {{{-b/(2a)}}}

So the x-coordinate of the vertex is {{{-b/(2a)=-(100)/(2(-2))=100/4=25}}}

So the value of x is 25

Now substitute 25 for x in 

{{{t+2x = 100}}}

{{{t+2(25) = 100}}}

{{{t+50 = 100}}}

{{{t=50}}}

So the two numbers are

first = 25
second = 50

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If you're taking calculus, you would do it by
taking the derivative of 

{{{y = 100x-2x^2}}}

{{{(dy)/(dx)=100-4x}}}

and set the derivative = 0

{{{100-4x=0}}}

{{{-4x=-100}}}

{{{x=25}}}

and as in the algebra way, {{{t=100-2x=100-2(25)=100-50=50}}} 

Edwin</pre>