Difference between revisions of "022 Exam 2 Sample B, Problem 8"

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(Created page with "<span class="exam"> Find the quantity that produces maximum profit, given demand function <math style="vertical-align: -15%">p = 70 - 3x</math> and cost function  <math...")
 
 
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Find the quantity that produces maximum profit, given demand function <math style="vertical-align: -15%">p = 70 - 3x</math> and cost function&thinsp; <math style="vertical-align: -8%">C = 120 - 30x + 2x^2.</math>
 
Find the quantity that produces maximum profit, given demand function <math style="vertical-align: -15%">p = 70 - 3x</math> and cost function&thinsp; <math style="vertical-align: -8%">C = 120 - 30x + 2x^2.</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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! Foundations: &nbsp;
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|-
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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>.
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Moreover, we have several important important functions:
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|-
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|
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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;<br>
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*<math style="vertical-align: -20%">R(x)</math>, the '''total revenue''' (or gross receipts) from producing <math style="vertical-align: 0%">x</math> units;<br>
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*<math style="vertical-align: -20%">P(x)</math>, the '''total profit''' from producing <math style="vertical-align: 0%">x</math> units.<br>
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|In particular, we have the relations
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|-
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::<math>P(x)=R(x)-C(x),</math>
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|and
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|-
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|
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::<math>R(x)=x\cdot p(x).</math>
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|-
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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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|}
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&nbsp;'''Solution:'''
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 1: &nbsp;
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|-
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|'''Find the Profit Function:''' We have
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|-
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|
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::<math>R(x)\,=\,x\cdot p(x)\,=\,x\cdot (70-3x)\,=\,70x-3x^2.</math>
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|-
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|From this,
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|-
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|
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::<math>P(x)\,=\,R(x)-C(x)\,=\,70x-3x^2- \left(120 - 30x + 2x^2 \right)\,=\,-120 + 100 x - 5 x^2 .</math>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 2: &nbsp;
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|-
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|'''Find the Maximum:''' The equation for profit at a given production level is
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|-
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|
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::<math>P(x)\,=\,-120 + 100 x - 5 x^2 .</math>
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|-
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|Applying our power rule to each term, we find
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|-
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|
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::<math>P'(x)\,=\,100-10x\,=\,10(10-x).</math>
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|The only root of this occurs at <math style="vertical-align: -3%">x=10</math>, and this is our production level to achieve maximum profit.
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|
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Final Answer: &nbsp;
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|-
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|Maximum profit occurs when we produce 10 items.
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|}
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[[022_Exam_2_Sample_B|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 17:32, 17 May 2015

Find the quantity that produces maximum profit, given demand function and cost function 

Foundations:  
Recall that the demand function, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x)} , relates the price per unit Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} to the number of units sold, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} .

Moreover, we have several important important functions:

  • Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C(x)} , the total cost to produce Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} units;
  • Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R(x)} , the total revenue (or gross receipts) from producing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} units;
  • Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(x)} , the total profit from producing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} units.
In particular, we have the relations
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(x)=R(x)-C(x),}
and
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R(x)=x\cdot p(x).}
Using these equations, we can find the maximizing production level by determining when the first derivative of profit is zero.

 Solution:

Step 1:  
Find the Profit Function: We have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R(x)\,=\,x\cdot p(x)\,=\,x\cdot (70-3x)\,=\,70x-3x^2.}
From this,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(x)\,=\,R(x)-C(x)\,=\,70x-3x^2- \left(120 - 30x + 2x^2 \right)\,=\,-120 + 100 x - 5 x^2 .}
Step 2:  
Find the Maximum: The equation for profit at a given production level is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(x)\,=\,-120 + 100 x - 5 x^2 .}
Applying our power rule to each term, we find
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P'(x)\,=\,100-10x\,=\,10(10-x).}
The only root of this occurs at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=10} , and this is our production level to achieve maximum profit.
Final Answer:  
Maximum profit occurs when we produce 10 items.


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