Question 957901: Please help me solve this question:
You will be constructing a rectangular deck against your bungalow, using 32 feet of railing and will leave a 4 foot gap in the railing for access to the stairs. Determine the dimensions that will maximize the area of the deck, and the maximum area.
What work I've done so far:
- The deck only has 3 sides (32 ft railing and a 4 ft gap)
36/3 = 3S/3
= 12 for Width of each side (therefore 12ft squared is the max area of the deck)
For max area I did
A= l x W
A= (12)X (12)
A= 144, Therefore 144 ft squared is the max area)
Gap for railing = 4ft.
A= max area - gap
A= 144-4
A= 140 ft squared.
Any help will be greatly appreciated!!
Found 2 solutions by macston, addingup:
Answer by macston(5194)   (Show Source):
You can put this solution on YOUR website! L=length=parallel to wall; W=width= perpendicular to wall
Available length=32ft rail + 4ft gap=36ft
L=36ft-2W
Area=LW=
This is a parabola: the maximum value will be at its vertex, which is where slope=0. The slope of a function is the first derivative, so:
dA=36-4W and we want the value where slope is 0:
36-4W=0
-4W=-36
W=9 The width will be 9 feet
L=36ft-2(W)=36ft-18 ft=18ft The length is 18 feet.
ANSWER 1: The dimensions for maximum area are 18 feet by 9 feet and the area is
A=LW
A=(18ft)(9ft)
A=162 square feet ANSWER 2: The maximum area is 162 square feet.
If this is not a calculus question, see Question 7782 for an alternative method.
NOTE: The gap in the railing does not change the area of the deck.
Answer by addingup(3677)  (Show Source):
You can put this solution on YOUR website! Total perimeter (3 sides only): 32+4= 36 (32 of railing + 4ft opening for stairs)
I’ll call the Length X and the Width Y:
Maximize A = area of the rectangle = (Length)(Width) = X*Y
given a perimeter of X+2Y= 36 (only 1 X, the other is against the bungalow-see my illustration)
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Now we need to solve for one of the variables. Since X+2Y= 36,
2Y= 36-X; Y= (36-X)/2; Y= 18- X/2
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Now that we solved for Y, we have that:
A= X*Y= X*(18- X/2)= 18X-X^2/2
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We want to maximize A:
A’= 18-X. The only time A’=0 is when X= 18.
So, 0<=X<=36. Why? Because we make it such that the only critical points of A are when X=0, 18, and 36, and the maximum area will be given by one of them.
Here we go:
At the critical number X= 0: A= 18(0) - (0)^2/2= 0 square feet – toss this answer out.
At the critical number X= 18: A= 18(18) – (18)^2/2 = 162
At the critical number X= 36: A= 18(36) – (36)^2/2 = 0 – toss this one out, too.
So our largest possible area is 162 square feet and the dimensions are: length (X)= 18 feet and width (Y)= 18 - X/2= 18 - 9= 9 feet.
Check: 18 + 2(9)= 36 feet, 32 of railing+4 feet opening on the railing for the stairs.
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