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

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(Created page with "Use differentials to approximate the change in profit given <math style="vertical-align: -5%">x = 10</math>  units and <math style="vertical-align: 0%">dx = 0.2</math>&...")
 
 
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Use differentials to approximate the change in profit given <math style="vertical-align: -5%">x = 10</math>&thinsp; units and <math style="vertical-align: 0%">dx = 0.2</math>&thinsp; units, where profit is given by  <math style="vertical-align: -23%">P(x) = -4x^2 + 90x - 128</math>.
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[[File:022_2_A_8.png|right|400px]]
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<span class="exam">Use differentials to approximate the change in profit given <math style="vertical-align: -5%">x = 10</math>&thinsp; units and <math style="vertical-align: 0%">dx = 0.2</math>&thinsp; units, where profit is given by  <math style="vertical-align: -23%">P(x) = -4x^2 + 90x - 128</math>.
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
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|A differential is a method of approximating a change,  
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|A differential is a method of linearly approximating the change of a function.  We use the derivative of the function at an initial point <math style="vertical-align: 0%">x_0</math> as the slope of a line, and use the standard relation
 
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::<math>\int x^n dn = \frac{x^{n+1}}{n+1} + C</math>
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::<math>m\,=\,\frac{\Delta y}{\Delta x},</math>
 
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|For setup of the problem we need to integrate the region between the x - axis, the curve, x = 1, and x = 4.
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|where <math style="vertical-align: -20%">\Delta y</math> represents the change in <math style="vertical-align: -20%">y</math> values, and <math style="vertical-align: 0%">\Delta x</math> represents the change in <math style="vertical-align: 0%">x</math> values. Due to the use of the derivative <math style="vertical-align: -22%">f'\left(x_0\right)</math> as the slope, we usually rewrite this using <math>dy</math> and <math style="vertical-align: 0%">dx</math> to indicate the relative changes. Thus,
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::<math>f'(x_0)\,=\,m\,=\,\frac{dy}{dx}.</math>
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|We can then rearrange this to find <math>dy=f'(x_0)\cdot dx.</math>
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!Step 1: &nbsp;
 
!Step 1: &nbsp;
 
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|First, we must find the derivative.  We have <math style="vertical-align: -23%">P'(x) = -8x + 90</math>.
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
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|We need the derivative at our initial point, or <math style="vertical-align: -15%">x_0 = 10</math>.  This is
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::<math>P'(x_0)\,=\,P'(10) \,=\, -8(10) + 90\,=\,10.</math>
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!Step 3: &nbsp;
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|Finally, we plug in the values to find
 
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::<math>dy\,=\,P'(x_0)\cdot dx\,=\,10\cdot 0.2\,=\,2.</math>
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|Note that if a teacher gives you units (thousands of dollars, dollars, cubits...), you should include them in your answer.
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Final Answer: &nbsp;
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::<math>dy\,=\,2.</math>
 
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[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]
 
[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 21:13, 15 May 2015

022 2 A 8.png

Use differentials to approximate the change in profit given   units and   units, where profit is given by .

Foundations:  
A differential is a method of linearly approximating the change of a function. We use the derivative of the function at an initial point as the slope of a line, and use the standard relation
where represents the change in values, and represents the change in values. Due to the use of the derivative as the slope, we usually rewrite this using and to indicate the relative changes. Thus,
We can then rearrange this to find

 Solution:

Step 1:  
First, we must find the derivative. We have .
Step 2:  
We need the derivative at our initial point, or . This is
Step 3:  
Finally, we plug in the values to find
Note that if a teacher gives you units (thousands of dollars, dollars, cubits...), you should include them in your answer.
Final Answer:  

Return to Sample Exam