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I solved a problem and am wondering if its right. Basically, a person has 150 feet of fence to put around a rectangular garden. A 10-foot opening is left on one side for a gate. I have to determine the length and width for the maximum area. Since one side has an opening of 10, the side parallel to it is also going to be 10. 150-10= 140, which I divided by 2. So the other two sides are each 70, making the maximum area 700 cm^2. Is this right?

2006-11-12 07:43:08 · 4 answers · asked by morgulis2003 3 in Science & Mathematics Mathematics

4 answers

Imagine the fence completley surrounding the garden:
then 2l+2w = 160 where l=length and w=width
so w = 80-l
also area, a=lw = 80l - l^2
for max area da/dl=0
da/dl = 80 - 2l
this is equal to zero when l=40 ft
so w=40 ft
and area = 1600 ft^2
(nb the fence would be 40ft long on 3 sides and 30ft long on one side leaving room for a 10ft gate - using 150ft of fence)

2006-11-12 07:49:53 · answer #1 · answered by ? 7 · 0 0

Area = l*w
Let x = width without gate
Then width with gate = x-10
The combined length of the two ends is thus 2x-10
This means that length = [150 -(2x-10)]/2= 75-x +5= 80-x
The area (A) then is (80-x)(x) = 80x -x^2
dA/dx= 80-2x
Setting this = 0 gives 80-2x=0 2x=80 x=40
length =80-x = 80-40 =40
Therefore the maximum area is a square with A=40^2 = 1600 sq ft. The opening for the fence is 10 ft so one end uses 30 ft of fencing and the other three sides all use 40 ft of fencing for a total of 30 +3*40 = 150 ft.

Calculus was needed for this problem to find the maximum.

You should also watch your units. The answer shoulld be in square feet, not square centimeters.

2006-11-12 16:22:44 · answer #2 · answered by ironduke8159 7 · 0 0

I don't think the opening is supposed to cover one full side of the fence; at least, it isn't apparent in the problem. The way I would set up this problem is like this:
area = length * width
circumference = 2 * length + 2 * width = 150 + 10
Solve the circumference equation for either length or width, and substitute it into the other equation. You will get a quadratic equation of the form area = (a bunch of stuff with either length or width in it) that looks like a parabola opening down on a graph. The vertex, which is the highest point on the curve, is the point at which area is maximized. The value of the y-coordinate is the max area, and the value of the x-coordinate is either the length or width that yields it.

2006-11-12 15:59:29 · answer #3 · answered by Anonymous · 0 0

70^2=4900

2006-11-12 15:47:50 · answer #4 · answered by      7 · 0 0

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