Unit 5

Applications of Derivatives

A. Maximum and Minimum Problems

Lesson 1: Numbers Problems and Geometric Applications


Geometric Applications

How would a farmer enclose a rectangular plot of land with a given amount of fencing to maximize the plot’s area? How would a manufacturer construct a box of maximum volume given a specific amount of material? These problems could be solved by trial and error, but there is greater efficiency with calculus.



Find the dimensions of a rectangle of maximum area with a perimeter of «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»28«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/mrow»«/mstyle»«/math». Determine the maximum area.

Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»l«/mi»«/math» and «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»w«/mi»«/math» represent the length and width of the rectangle and let «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mi»P«/mi»«/mstyle»«/math» be the perimeter.

Determine the length «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mi»l«/mi»«/mstyle»«/math» in terms of width «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mi»w«/mi»«/mstyle»«/math» using the perimeter formula for a rectangle.

«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»P«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mi mathvariant=¨normal¨»(«/mi»«mi»l«/mi»«mo»+«/mo»«mi»w«/mi»«mi mathvariant=¨normal¨»)«/mi»«/mtd»«/mtr»«mtr»«mtd»«mn»28«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mi mathvariant=¨normal¨»(«/mi»«mi»l«/mi»«mo»+«/mo»«mi»w«/mi»«mi mathvariant=¨normal¨»)«/mi»«/mtd»«/mtr»«mtr»«mtd»«mn»14«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»l«/mi»«mo»+«/mo»«mi»w«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»l«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»14«/mn»«mo»§#8722;«/mo»«mi»w«/mi»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»

Let «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mi»A«/mi»«/mstyle»«/math» be the area of the rectangle.  

 «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»A«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»l«/mi»«mi»w«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi mathvariant=¨normal¨»(«/mi»«mn»14«/mn»«mo»§#8722;«/mo»«mi»w«/mi»«mi mathvariant=¨normal¨»)«/mi»«mi»w«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»14«/mn»«mi»w«/mi»«mo»§#8722;«/mo»«msup»«mi»w«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»

Determine the derivative of the function, and set it equal to zero to find the critical number(s).

«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»w«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»14«/mn»«mo»-«/mo»«mn»2«/mn»«mi»w«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd/»«mtd/»«/mtr»«mtr»«mtd»«mn»0«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»14«/mn»«mo»-«/mo»«mn»2«/mn»«mi»w«/mi»«/mtd»«/mtr»«mtr»«mtd»«mn»2«/mn»«mi»w«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»14«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»w«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»7«/mn»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»

Note:  The notation «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mfrac»«mrow»«mi»d«/mi»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»w«/mi»«/mrow»«/mfrac»«/mstyle»«/math» is used because the derivative of «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mi»A«/mi»«/mstyle»«/math» is being determined with respect to the variable «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mi»w«/mi»«/mstyle»«/math».

Solve for «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mi»l«/mi»«/mstyle»«/math» when «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»w«/mi»«mo»=«/mo»«mn»7«/mn»«/mrow»«/mstyle»«/math».
        
«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»l«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»14«/mn»«mo»§#8722;«/mo»«mi»w«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»14«/mn»«mo»§#8722;«/mo»«mn»7«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»7«/mn»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»

«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»w«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»14«/mn»«mo»§#8722;«/mo»«mn»2«/mn»«mi»w«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«msup»«mi»d«/mi»«mn»2«/mn»«/msup»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«msup»«mi»w«/mi»«mn»2«/mn»«/msup»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»-«/mo»«mn»2«/mn»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»

By the second derivative test, the function is at a maximum because the second derivative is negative for all «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mi»x«/mi»«/mstyle»«/math» (concave down).

The dimensions of the rectangle with maximum area are «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»7«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/mrow»«/mstyle»«/math» by «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»7«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/mrow»«/mstyle»«/math».  The maximum area is «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mn»49«/mn»«mo»§#160;«/mo»«msup»«mi»cm«/mi»«mn»2«/mn»«/msup»«/mstyle»«/math».
Watch the video Steps to Solving Maximum and Minimum Problems to see a real-life application of how to optimize the volume of a box.