Difference between revisions of "Math 22 Optimization Problems"

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==Solving Optimization Problems==
 
==Solving Optimization Problems==
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'''Find Maximum Area''': Find the length and width of a rectangle that has 80 meters perimeter and a maximum area.
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'''1) Maximum Area''': Find the length and width of a rectangle that has 80 meters perimeter and a maximum area.
  
 
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{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
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|}
 
|}
  
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'''2) Maximum Volume''' A rectangular solid with a square base has a surface area of <math>337.5</math> square centimeters. Find the dimensions that yield the maximum volume.
  
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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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|Let <math>a</math> be the length of the one side of the square base in centimeter.
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|-
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|and <math>h</math> be the height of the solid in centimeter.
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|-
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|Then, the surface area <math>S=2a^2+4ah=337.5</math>, so <math>h=\frac{337.5-2a^2}{4a}</math>
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|-
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|Volume <math>V=a^2h=a^2(\frac{337.5-2a^2}{4a})=\frac{1}{4}a(337.5-a^2)=\frac{337.5a}{4}-\frac{a^3}{4}</math>
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|-
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|<math>V'=\frac{337.5}{4}-\frac{3}{4}a^2=0</math>, then <math>a^2=\frac{225}{4}</math>, so <math>a=\pm\frac{25}{2}=\frac{25}{2}</math> since <math>a</math> is positive.
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|-
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|Hence, <math>h=\frac{337.5-2a^2}{4a}=h=\frac{337.5-2(\frac{25}{2})^2}{4\frac{25}{2}}=\frac{1}{2}</math>
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|-
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|Therefore, the dimensions that yield the maximum value is <math>a=\frac{25}{2}</math> and <math>h=\frac{1}{2}</math>
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|}
  
  

Revision as of 08:14, 1 August 2020

Solving Optimization Problems

1) Maximum Area: Find the length and width of a rectangle that has 80 meters perimeter and a maximum area.

Solution:  
Let 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 l} be the length of the rectangle in meter.
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 w} be the width of the rectangle in meter.
Then, the perimeter 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=2(l+w)=80} , 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 l+w=40} , 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 l=40-w}
Area 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 A=l.w=(40-w)w=40w-w^2}
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 A'=40-2w=0} , 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 w=20} , 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 l=40-w=40-20=20}
Therefore, 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 l=w=20}

2) Maximum Volume A rectangular solid with a square base has a surface area of 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 337.5} square centimeters. Find the dimensions that yield the maximum volume.

Solution:  
Let 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 a} be the length of the one side of the square base in centimeter.
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 h} be the height of the solid in centimeter.
Then, the surface area 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 S=2a^2+4ah=337.5} , 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 h=\frac{337.5-2a^2}{4a}}
Volume 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 V=a^2h=a^2(\frac{337.5-2a^2}{4a})=\frac{1}{4}a(337.5-a^2)=\frac{337.5a}{4}-\frac{a^3}{4}}
, 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 a^2=\frac{225}{4}} , 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 a=\pm\frac{25}{2}=\frac{25}{2}} 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 a} is positive.
Hence, 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 h=\frac{337.5-2a^2}{4a}=h=\frac{337.5-2(\frac{25}{2})^2}{4\frac{25}{2}}=\frac{1}{2}}
Therefore, the dimensions that yield the maximum value 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 a=\frac{25}{2}} 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 h=\frac{1}{2}}


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