Question 245998: A carpenter is building a rectangular room with a fixed perimeter of 240 ft. What dimensions would yeild the maximum area? what is the maximum area?
Found 2 solutions by ripplemehta, oberobic:
Answer by ripplemehta(3)   (Show Source):
You can put this solution on YOUR website! perimeter of the rectangulatr room=240ft
perimeter of a rectangle=2*(l+b)
so 240=2*(l+b)so
l+b=120
so if l=x
b=120-x
area=x*(120-x)
A=x(120-x)
for maximum A dA/dx=0
dA/dx=d/dx(120x-x^2)
0=120-2x
x=60ft
length=60ft
breadth=120-60=60ft answer
Answer by oberobic(2304)  (Show Source):
You can put this solution on YOUR website! P = 240 = 2L + 2W
A = L*W
.
By theorem (or from calculus), we know the maximum area is a square, so: L=W.
.
P = 2L + 2W = 4L = 240
L = 60
.
So a 60 by 60 room is 3600 sq ft in area.
.
We can test the area by testing another room with dimensions close to 60 by 60 and P = 240.
.
Let's try 59 by 61. P = 2(59)+2(61) = 118+122 = 240. OK.
A = 59 * 61 = 3599, which is less than 3600.
.
Let's try 58 by 62. P = 2(58) + 2(62) = 116 + 124 = 240. OK
A = 58 * 63 = 3596, which is less than 3600 and less than 3599.
.
Let's try 57 by 63. P = 240. OK
A = 57 * 63 = 3591, which is less than 3600 and less than 3599 and less than 3596. The area is decreasing as the values get further apart.
.
How about 1 by 119? P = 2(1) + 2(119) = 2 + 238 = 240.
A = 1*119 = 119, which is way less...
.
Another way to look at this problem is the define the perimeter = 2C, where C = L+W.
.
That way L = C-W, so the area will be (C-W) * W = CW - W^2.
.
The maximum value of CW - W^2 is C^2/4, which occurs at W = C/2.
.
So, if W = C/2, and C=L+W, then L=C/2, so we have a square.
|
|
|
