Difference between revisions of "022 Exam 2 Sample A, Problem 10"

Revision as of 07:14, 16 May 2015

Use calculus to set up and solve the word problem: Find the length and width of a rectangle that has a perimeter of 48 meters and maximum area.

Foundations:
As with all word problems, start with a picture. Using the variables $x$ and $y$ as shown in the image, we need to remember the equations of a rectangle for perimeter:
$P\,=\,2x+2y,$
and for area:
$A\,=\,xy.$
Since we want to maximize area, 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 maximum area.

Solution:

Step 1:
Express one variable in terms of the other: Since we know that the perimeter is to be 48 meters and $P\,=\,2x+2y$ , we can solve for $y$ in terms of $x$ . Since
$48\,=\,P\,=\,2x+2y,$
we find that $y=24-x.$

Step 2:
Find an expression for area in terms of one variable: Now, we can use the substitution from part 1 to find
$A(x)\,=\,xy\,=\,x(24-x)\,=\,24x-x^{2}.$

Step 3:
Find the derivative and its roots: Since $A(x)=24x-x^{2}$ , we have
$A'(x)\,=\,24-2x\,=\,2(12-x).$
This derivative is zero precisely when $x=y=12$ , and these are the values that will maximize area. Also, don't forget the units - meters!

Final Answer:
The area is maximized when both the length and width are 12 meters.

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|As with all word problems, start with a picture.  Using the variables <math style="vertical-align: 0%">x</math> and <math style="vertical-align: 0%">y</math> as shown in the image, we need to remember the equations for perimeter:
+|As with all word problems, 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 perimeter:
 |-
 |
 ::<math>P\,=\,2x+2y,</math>
 |-
-|and that for area:
+|and for area:
 |-
 |
 ::<math>A\,=\,xy.</math>
 |-
 |Since we want to maximize area, 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 maximum area.
 |}
@@ Line 50: / Line 50: @@
 ::<math>A'(x)\,=\,24-2x\,=\,2(12-x).</math>
 |-
-|This derivative is zero precisely when <math style="vertical-align: -20%">x=y=12</math>, and these are the values that will maximize area.
+|This derivative is zero precisely when <math style="vertical-align: -20%">x=y=12</math>, and these are the values that will maximize area.  Also, don't forget the units - meters!
 |}