Question 129106
a)


The revenue can be found by this general formula:


Revenue=Number of units sold*price of each unit




So in our case, {{{R(x)=x*P(x)}}}



{{{R(x)=x*P(x)}}} Start with the given equation



{{{R(x)=x*(100-0.05x)}}} Plug in {{{P(x) = 100 – 0.05x}}}



{{{R(x)=100x-0.05x^2}}} Distribute



{{{R(x)=-0.05x^2+100x }}} Rearrange the terms



So the revenue function is 


{{{R(x)=-0.05x^2+100x }}}




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b)


The profit can be found by:


Profit=Revenue-Cost


So 


{{{P(x)=R(x)-C(x)}}} 



{{{P(x)=(-0.05x^2+100x)-(4000 + 60x-0.01x^2)}}} Plug in {{{R(x)=-0.05x^2+100x }}} and {{{C(x)=4000 + 60x – 0.01x^2}}} 



{{{P(x)=-0.05x^2+100x-4000-60x + 0.01x^2}}} Distribute the negative



{{{P(x)=-0.04x^2+40x-4000}}} Combine like terms



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To find the max profit, we need to graph the profit function {{{P(x)=-0.04x^2+40x-4000}}}



{{{ graph( 500, 500,-740,1260,-10523,7649,-0.04x^2+40x-4000)}}}



And by using the calculators vertex function, we find that the vertex occurs at {{{x=500}}}



Now plug in {{{x=500}}} into {{{P(x)=-0.04x^2+40x-4000}}}



{{{P(x)=-0.04(500)^2+40(500)-4000}}}


{{{P(x)=6000}}} Simplify



So the maximum profit is $6,000