SOLUTION: A carpenter is building a rectangular room with a fixed perimeter of 304 ft. What dimensions would yield the maximum area? What is the maximum area?
The length that would yield
Question 395574: A carpenter is building a rectangular room with a fixed perimeter of 304 ft. What dimensions would yield the maximum area? What is the maximum area?
The length that would yield the maximum area is ___ ft. Found 2 solutions by solver91311, richard1234:Answer by solver91311(24713) (Show Source):
The dimensions of a rectangle that give the maximum area for a given perimeter are , that is to say it is a square whose side measure is one fourth of the given perimeter.
Proof:
Let represent the length of the rectangle and let represent the width.
The perimeter of a rectangle that measures by is .
A little algebra:
The area of the rectangle is given by .
Substituting:
Which can be re-written:
From here you can either use the formula for the x-coordinate of the vertex of:
Which is:
since we know that with a negative lead coefficient, the parbola opens down and thus the vertex is a function maximum.
meaning that the area is maximum when the width (and therefore the length as well) is one-fourth of the perimeter.
Or you can set the first derivative equal to zero:
And then evaluate the second derivative at :
Since in the domain of , the value of is a local maximum.
John
My calculator said it, I believe it, that settles it
You can put this solution on YOUR website! There are multiple ways to prove that the maximum area is a square. You could let A = w*(152 - w), then take the derivative or find the vertex, as the other tutor did.
I like the classic AM-GM inequality: AM-GM (arithmetic mean - geometric mean inequality) says that for positive real numbers , the arithmetic mean of all 's is greater than or equal to the geometric mean. If we let w and 152-w be the sides,
--> the area is less than or equal to 76^2. AM-GM says that the equality case occurs if and only if all the 's are equal, or w = 304-w --> w = 76, area = 76^2.
This is a slightly unconventional way, but it's the easiest way to prove the similar problem that says that the largest possible volume of a rectangular solid of fixed surface area occurs when the solid is a cube.
