227k views
5 votes
Find the dimensions of a rectangle with perimeter 112 m whose area is as large as possible. step 1 if a rectangle has dimensions x and y, then we must maximize the area a = xy. since the perimeter is 2x + 2y = 112, then y = 56

User Golinmarq
by
8.1k points

1 Answer

2 votes


Please use the solution below:
Let P = perimeter, A = area
As provided above, a = xy and
We know that the formula to solve for the perimeter of a rectangle is P = 2x + 2y. Using the given 112m, we can solve the perimeter using the formula:
112 = 2x + 2y
56 = x + y

x = 56-y or y = 56-x
Let's solve the perimeter in terms of y using the formula below:
A = (56-y)(y)
Find the derivative of A = 56-y^2 to get the value of y.
dA/dy = 56-2y = 0
y = 56/2
y = 28
To find X, substitute the value of y in the equation x = 56 - y.
x = 56 - 28
Therefore, x = 28.
We can conclude that the figure is not a rectangle but a square.
User Brian Powell
by
7.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories