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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|-
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|In particular, we have the relations
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|-
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|
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::<math>P(x)=R(x)-C(x),</math>
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|-
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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 (90-3x)\,=\,90x-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)\,=\,90x-3x^2- \left(200-30x+x^2 \right)\,=\,120x-4x^2-200 .</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 marginal revenue is
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|-
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|
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::<math>P(x)\,=\,120x-4x^2-200 .</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)\,=\,120-8x\,=\,8(15-x).</math>
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|-
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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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|-
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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 15 items.
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|}
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[[022_Exam_2_Sample_B|'''<u>Return to Sample Exam</u>''']]

Revision as of 17:41, 15 May 2015

Find the quantity that produces maximum profit, given demand function and cost 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 C = 120 - 30x + 2x^2.}

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 (90-3x)\,=\,90x-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)\,=\,90x-3x^2- \left(200-30x+x^2 \right)\,=\,120x-4x^2-200 .}
Step 2:  
Find the Maximum: The equation for marginal revenue 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)\,=\,120x-4x^2-200 .}
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)\,=\,120-8x\,=\,8(15-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=15} , and this is our production level to achieve maximum profit.
Final Answer:  
Maximum profit occurs when we produce 15 items.


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