Difference between revisions of "022 Exam 2 Sample B, Problem 10"
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<span class="exam">'''Use calculus to set up and solve the word problem:''' | <span class="exam">'''Use calculus to set up and solve the word problem:''' | ||
A fence is to be built to enclose a rectangular region of 480 square feet. The fencing material along three sides cost $2 per foot. The fencing material along the 4<sup>th</sup> side costs $6 per foot. Find the most economical dimensions of the region (that is, minimize the cost). | A fence is to be built to enclose a rectangular region of 480 square feet. The fencing material along three sides cost $2 per foot. The fencing material along the 4<sup>th</sup> side costs $6 per foot. Find the most economical dimensions of the region (that is, minimize the cost). | ||
+ | |||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | !Foundations: | ||
+ | |- | ||
+ | |As with all geometric word problems, it helps to start with a picture. Using the variables <math style="vertical-align: 0%">x</math> and <math style="vertical-align: -20%">y</math> as shown in the image, we need to remember the equations of a rectangle for area: | ||
+ | |- | ||
+ | | | ||
+ | ::<math>A\,=\,xy.</math> | ||
+ | |- | ||
+ | |However, we need to construct a new function to describe cost: | ||
+ | |- | ||
+ | | | ||
+ | ::<math>C\,=\,(2+6)x+(2+2)y\,=\,8x+4y.</math> | ||
+ | |- | ||
+ | |Since we want to minimize cost, we will have to rewrite it as a function of a single variable, and then find when the first derivative is zero. From this, we will find the dimensions which provide the minimum cost. | ||
+ | |} | ||
+ | |||
+ | '''Solution:''' | ||
+ | |||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | !Step 1: | ||
+ | |- | ||
+ | |'''Express one variable in terms of the other:''' Since we know that the area is 480 square feet and <math style="vertical-align: -20%">A\,=\,xy</math>, we can solve for <math style="vertical-align: -15%">y</math> in terms of <math style="vertical-align: 0%">x</math>. Since <math style="vertical-align: -20%">480\,=\,xy</math>, we find that <math style="vertical-align: -20%">y=480/x</math>. | ||
+ | |} | ||
+ | |||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | !Step 2: | ||
+ | |- | ||
+ | |'''Find an expression for cost in terms of one variable:''' Now, we can use the substitution from part 1 to find | ||
+ | |- | ||
+ | | | ||
+ | ::<math>C(x)\,=\,8x+4y\,=\,8x+4\cdot \frac{480}{x}\,=\,8x+\frac{1920}{x}.</math> | ||
+ | |} | ||
+ | |||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | !Step 3: | ||
+ | |- | ||
+ | |'''Find the derivative and its roots:''' We can apply the power rule term-by-term to find | ||
+ | |- | ||
+ | | | ||
+ | ::<math>C'(x)\,=\,8-\frac{1920}{x^2}\,=\,8\left(1-\frac{240}{x^2}\right).</math> | ||
+ | |- | ||
+ | |This derivative is zero precisely when <math style="vertical-align: -10%">x=4\sqrt{15}</math>, which occurs when <math style="vertical-align: -20%">y=8\sqrt{15}</math>, and these are the values that will minimize cost. Also, don't forget the units - feet! | ||
+ | |} | ||
+ | |||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | !Final Answer: | ||
+ | |- | ||
+ | |The cost is minimized when the dimensions are <math style="vertical-align: -5%">8\sqrt{15}</math> feet by <math style="vertical-align: -5%">4\sqrt{15}</math> feet. Note that the side with the most expensive fencing is the shorter one. | ||
+ | |} | ||
+ | |||
+ | [[022_Exam_2_Sample_B|'''<u>Return to Sample Exam</u>''']] |
Revision as of 11:23, 17 May 2015
Use calculus to set up and solve the word problem: A fence is to be built to enclose a rectangular region of 480 square feet. The fencing material along three sides cost $2 per foot. The fencing material along the 4th side costs $6 per foot. Find the most economical dimensions of the region (that is, minimize the cost).
Foundations: |
---|
As with all geometric word problems, it helps to start with a picture. Using the variables and as shown in the image, we need to remember the equations of a rectangle for area: |
|
However, we need to construct a new function to describe cost: |
|
Since we want to minimize cost, we will have to rewrite it as a function of a single variable, and then find when the first derivative is zero. From this, we will find the dimensions which provide the minimum cost. |
Solution:
Step 1: |
---|
Express one variable in terms of the other: Since we know that the area is 480 square feet and , we can solve for in terms of . Since , we find that . |
Step 2: |
---|
Find an expression for cost in terms of one variable: Now, we can use the substitution from part 1 to find |
|
Step 3: |
---|
Find the derivative and its roots: We can apply the power rule term-by-term to find |
|
This derivative is zero precisely when , which occurs when , and these are the values that will minimize cost. Also, don't forget the units - feet! |
Final Answer: |
---|
The cost is minimized when the dimensions are feet by feet. Note that the side with the most expensive fencing is the shorter one. |