Difference between revisions of "Math 22 Business and Economics Applications"

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This page is under construction.
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==Optimization in Business and Economics==
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'''1)''' Find the number of units <math>x</math> that minimizes the average cost per unit <math>\overline{C}</math> when <math>C=2x^2+348x+7200</math>
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{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
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!Solution: &nbsp;
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|-
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|Notice: <math>\overline{C}=\frac{C}{x}=\frac{2x^2+348x+7200}{x}=2x+348+\frac{7200}{x}</math>
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|Then, <math>\overline{C} '=2-\frac{7200}{x^2}=0</math>, so <math>x^2=3600</math>, so <math>x=\pm\sqrt{3600}=\pm 60=60</math> since <math>x</math> is positive.
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'''2)''' Find the price that will maximize profit for the demand and cost functions, where <math>p</math> is the price, <math>x</math> is the number of units, and <math>C</math> is the cost. Given the demand function <math>p(x)=90-x</math> and the cost function <math>C(x)=100+30x</math>.
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{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
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!Solution: &nbsp;
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|Notice: The revenue function <math>R(x)=x\cdot p(x)=x(90-x)=90x-x^2</math>
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|The Profit function is <math>P(x)=R(x)-C(x)=90x-x^2-(100+30x)=90x-x^2-100-30x=-x^2+60x-100</math>
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|Then, <math>P'(x)=-2x+60=0</math>, so <math>x=30</math>
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|So, <math>p(30)=90-30=60</math>
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|Therefore, the price is <math>\$ 60</math> a unit will maximize the profit.
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[[Math_22| '''Return to Topics Page''']]
 
[[Math_22| '''Return to Topics Page''']]
  
 
'''This page were made by [[Contributors|Tri Phan]]'''
 
'''This page were made by [[Contributors|Tri Phan]]'''

Latest revision as of 07:51, 2 August 2020

Optimization in Business and Economics

1) Find the number of units that minimizes the average cost per unit when

Solution:  
Notice:
Then, , so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x^{2}=3600} , so 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=\pm\sqrt{3600}=\pm 60=60} since 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} is positive.

2) Find the price that will maximize profit for the demand and cost functions, where 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} is the price, 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} is the number of units, 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 C} is the cost. Given 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)=90-x} and the 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(x)=100+30x} .

Solution:  
Notice: The revenue 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 R(x)=x\cdot p(x)=x(90-x)=90x-x^2}
The Profit function 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)=R(x)-C(x)=90x-x^2-(100+30x)=90x-x^2-100-30x=-x^2+60x-100}
Then, 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)=-2x+60=0} , so 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=30}
So, 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(30)=90-30=60}
Therefore, the price 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 \$ 60} a unit will maximize the profit.




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