Question 1205103


{{{R(x)= x(4-0.0001x)}}}

{{{R(x)= 4x-0.0001x^2}}}

using calculus:
first derivate

{{{(d/dx)(4x-0.0001x^2)=4-2*0.0001x=4 - 0.0002x}}}

equal it to zero

{{{4 - 0.0002x=0}}}

{{{4 =0.0002x}}}

{{{x=4/0.0002}}}

{{{x=20000}}}

plug it in given equation 

{{{R(x)= 4*(20000)-0.0001*(20000)^2=40000}}}

max revenue is {{{40000}}}


or, solve it using algebra:

{{{R(x)= 4x-0.0001x^2}}}

The revenue equation is a quadratic, so its graph is a parabola.  Since the coefficient of the {{{x^2}}} term is negative ({{{-0.0001}}}), it's an inverted parabola with the vertex at the top.  
The vertex will thus be the {{{maximum}}} revenue.  
To find the vertex, convert {{{R}}} to the vertex form. 

{{{R = a(x-h)^2+k}}} where ({{{h}}},{{{k}}}) is the location of the vertex.  

Convert to the vertex form by completing the square:

{{{R(x)= 4x-0.0001x^2}}}

{{{R(x)=-0.0001( 4x/-0.0001+x^2)}}}

{{{R(x)=-0.0001( -40000x+x^2)}}}

{{{R(x)=-0.0001( b^2-40000x+x^2)-(-0.0001b^2)}}}...{{{b=40000/2=20000}}}

  rearrange the terms in parenthesis

{{{R(x)=-0.0001( x^2-40000x+20000^2)-(-0.0001*20000^2)}}}

{{{R(x)=-0.0001( x-20000)^2-(-40000)}}}

{{{R(x)=-0.0001( x-20000)^2+40000}}}

{{{h=20000}}}
{{{k=40000}}}

vertex is at ({{{20000}}},{{{40000}}})=>the maximum revenue is {{{40000}}}