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Tiffany is constructing a fence around a rectangular tennis court. She must use exactly 300 feet of fencing. The fence must enclose all four sides of the court. Regulation states that the length of the fence enclosure must be at least 80 feet and the width must be at least 40 feet. Tiffany wants the area enclosed by the fence to be as large as possible in order to accommodate benches and storage space. What is the optimal area, in square feet?

 Dec 26, 2018
 #1
avatar+100791 
+3

Tiffany is constructing a fence around a rectangular tennis court. She must use exactly 300 feet of fencing. The fence must enclose all four sides of the court. Regulation states that the length of the fence enclosure must be at least 80 feet and the width must be at least 40 feet. Tiffany wants the area enclosed by the fence to be as large as possible in order to accommodate benches and storage space. What is the optimal area, in square feet?

 

I can see straight off that the answer is going to be 80 by 70feet.

This is because the ratio of area to perimeter is highest for a circle so the closer that a 2D shape is to a circle the greast the area to perimeter ratio will be.  If it was a square then all sides would be 75 feet. The closest we are allowed to get to that is 80feet for the length. So a length of 80 feet will give max area.

 

BUT I expect you are supposed to do it via calculus.

 

Let the fence be    x+80  by  y+40   where x and y are greater or equal to zero.

so    

x+80+x+80+y+40+y+40=300

x+80+y+40=150

x+y=30

y=30-x

So the dimensions are  x+80  and  30-x+40 = 70-x

    

 


\(Area =(x+80)(70-x)\\ Area=-x^2+70x-80x+5600\\ Area=-(x^2+10x-5600)\\ \)

 

This is a concave down parabola so any turning point will be a maximum.

Find turning point     A'=0

 

\(A'=-2x-10=0\\ x=-5\\ \text{but x must be greater or equal to zero so the maximum area will be when x=0}\)

 

So maximum area occurs when the dimensions are 80feet  by 70 feet.

 

\(80*70=5600\;\; feet \;\;squared\)

.
 Dec 26, 2018
 #2
avatar+100471 
+2

Thanks, Melody

 

This problem can also be solved by a graphical method

 

Call x  the width and y the length

 

We have these constraints

 

 x ≥ 80

y  ≥ 40

Using these the area, xy,  must be ≥ 3200

 

And we have the final  constraint that  2x + 2y = 300  ⇒  x + y = 150

 

The graph shows that the point (80,70) is the  optimum solution....just as Melody found!!!

 

 

 

cool cool cool

 Dec 26, 2018
 #3
avatar+100791 
+1

Thanks Chis,

Yes that is another interesting answer :)

Melody  Dec 26, 2018

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