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A regulation NFL playing field of length
x and width y has a perimeter of (1040/3) yards.
(A) Show that the width of the rectangle is y = [(520/3) - x]
and its area is A = x[(520/3) - x)]
(B) From the graph, estimate the dimensions
of the rectangle that yield a maximum area.
1. Can someone set up part (A)? I will do the math. I know that
P = 2L + 2W is used here somehow, right?
2. For part (B), how do I use the graph of A = x[(520/3) - x)]
to estimate the dimensions of the rectangle that yield a maximum area.
Last edited by mathland (2021-04-26 14:34:19)
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Yes, that perimeter formula is the place to start. Put in y and x and the known value for P. then it's just a case of making y the subject of the equation.
Then area = xy so you can use that equation for y to get area in terms of x only.
Bob
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Yes, that perimeter formula is the place to start. Put in y and x and the known value for P. then it's just a case of making y the subject of the equation.
Then area = xy so you can use that equation for y to get area in terms of x only.
Bob
What about part B?
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Bob wrote:Yes, that perimeter formula is the place to start. Put in y and x and the known value for P. then it's just a case of making y the subject of the equation.
Then area = xy so you can use that equation for y to get area in terms of x only.
Bob
What about part B?
This is a classic optimization problem - just graph A(x) and find its local maximum in a "reasonable" domain of x(i.e, from 0 to 520/3, as area can't be negative.)
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mathland wrote:Bob wrote:Yes, that perimeter formula is the place to start. Put in y and x and the known value for P. then it's just a case of making y the subject of the equation.
Then area = xy so you can use that equation for y to get area in terms of x only.
Bob
What about part B?
This is a classic optimization problem - just graph A(x) and find its local maximum in a "reasonable" domain of x(i.e, from 0 to 520/3, as area can't be negative.)
I will use Desmos to graph the area function given to answer part B. I'll come back to this problem if need be.
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