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2015-05-15T14:09:47Z

MathAdmin: Created page with "Find the quantity that produces maximum profit, given the demand function $p\,=\,90-3x$ and cost function
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Created page with "Find the quantity that produces maximum profit, given the demand function <math style="vertical-align: -15%">p\,=\,90-3x</math> and cost function <math styl..."

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Find the quantity that produces maximum profit, given the demand function <math style="vertical-align: -15%">p\,=\,90-3x</math> and cost function <math style="vertical-align: -5%">C\,=\,200-30x+x^2</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations:  
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|Recall that the '''demand function''', <math style="vertical-align: -25%">p(x)</math>, relates the price per unit <math style="vertical-align: -17%">p</math> to the number of units sold, <math style="vertical-align: 0%">x</math>.
Moreover, we have several important important functions:
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<math style="vertical-align: -20%">C(x)</math>, the '''total cost''' to produce <math style="vertical-align: 0%">x</math> units; 
<math style="vertical-align: -20%">R(x)</math>, the '''total revenue''' (or gross receipts) from producing <math style="vertical-align: 0%">x</math> units; 
*<math style="vertical-align: -20%">P(x)</math>, the '''total profit''' from producing <math style="vertical-align: 0%">x</math> units. 
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|In particular, we have the relations
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::<math>P(x)=R(x)-C(x),</math>
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|and
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::<math>R(x)=x\cdot p(x).</math>
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|Using these equations, we can find the maximizing production level by determining when the first derivative of profit is zero.

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 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|'''Find the Profit Function:''' We have
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::<math>R(x)\,=\,x\cdot p(x)\,=\,x\cdot (90-3x)\,=\,90x-3x^2.</math>
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|From this,
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::<math>P(x)\,=\,R(x)-C(x)\,=\,90x-3x^2- \left(200-30x+x^2 \right)\,=\,120x-4x^2-200 .</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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|'''Find the Maximum:''' The equation for marginal revenue is

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::<math>P(x)\,=\,120x-4x^2-200 .</math>
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|Applying our power rule to each term, we find
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::<math>P'(x)\,=\,120-8x\,=\,8(15-x).</math>
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|The only root of this occurs at <math style="vertical-align: -5%">x=15</math>, and this is our production level to achieve maximum profit.
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|Maximum profit occurs when we produce 15 items.
|}

[[022_Exam_1_Sample_A|'''Return to Sample Exam''']]

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