Difference between revisions of "022 Exam 2 Sample A, Problem 8"
Jump to navigation
Jump to search
Line 1:  Line 1:  
+  [[File:022_2_A_8.pngright400px]]  
<span class="exam">Use differentials to approximate the change in profit given <math style="verticalalign: 5%">x = 10</math>  units and <math style="verticalalign: 0%">dx = 0.2</math>  units, where profit is given by <math style="verticalalign: 23%">P(x) = 4x^2 + 90x  128</math>.  <span class="exam">Use differentials to approximate the change in profit given <math style="verticalalign: 5%">x = 10</math>  units and <math style="verticalalign: 0%">dx = 0.2</math>  units, where profit is given by <math style="verticalalign: 23%">P(x) = 4x^2 + 90x  128</math>.  
Line 24:  Line 25:  
!Step 1:  !Step 1:  
    
−    +  First, we must find the derivative. We have <math style="verticalalign: 23%">P'(x) = 8x + 90</math>. 
}  }  
{ class="mwcollapsible mwcollapsed" style = "textalign:left;"  { class="mwcollapsible mwcollapsed" style = "textalign:left;"  
!Step 2:  !Step 2:  
+    
+  We need the derivative at our initial point, or <math style="verticalalign: 15%">x_0 = 10</math>. This is  
    
    
+  ::<math>P'(x_0)\,=\,P'(10) \,=\, 8(10) + 90\,=\,10.</math>  
+  }  
+  { class="mwcollapsible mwcollapsed" style = "textalign:left;"  
+  !Step 3:  
+    
+  Finally, we plug in the values to find  
+    
+    
+  ::<math>dy\,=\,P'(x_0)\cdot dx\,=\,10\cdot 0.2\,=\,2.</math>  
+    
+  Note that if a teacher gives you units (thousands of dollars, dollars, cubits...), you should include them in your answer.  
+  }  
+  { class="mwcollapsible mwcollapsed" style = "textalign:left;"  
+  !Final Answer:  
+    
+    
+  ::<math>dy\,=\,2.</math>  
}  }  
−  
[[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
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: 

