Question 476978: A farmer wants to make a vegetable garden such that length is 3 meters longer than the width. If the farmer has 70 meters of fencing material and he wants his garden to have an of at least 180 square meters, what are the smallest dimensions that the garden could have? What would the largest that it could have? in Complete Solution
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! L = length
W = width
L = W + 3
A = area
P = perimeter
P = 2L + 2W
A = L*W
The area has to be greater than 180 square meters.
This means that:
L*W >= 180
The perimeter has to be smaller than or equal to 70 meters.
This means that:
2L + 2W <= 70
the values for L and W that you are looking for will have to be able to satisfy both equations simultaneously.
Since L = W+3, we can substitute for L in these equation to get:
W^2 + 3W >= 180
4W + 6 <= 70
From the second equation, we get:
W <= 16
From the first equation, we get:
(W+15)*(W-12) >= 0
This means that (W+15) and (W-12) either both have to be positive or both have to be negative.
If both are positive, then W would have to be >= 12 because W-12 would be equal to 0 or less than 0 otherwise.
If both are negative, then W would have to be <= -15 because W + 15 would be positive otherwise.
So we have a situation where:
W >= 12 or W <= -15.
Since W could never be negative, then we have a situation where:
W would have to be greater than or equal to 12.
From the perimeter equation W needs to be less than or equal to 16.
From the area equation W needs to be greater than or equal to 12.
That means:
12 <= W <= 16
There are 6 possible values for W.
They are:
12
13
14
15
16
Since L is equal to W + 3, then there are 4 possible values of L corresponding to the 4 possible values of W.
The value pairs are:
W L
12 15
13 16
14 17
15 18
16 19
These value pairs would result in a Perimeter and an Area of:
W L P A
12 15 54 180
13 16 58 208
14 17 62 238
15 18 66 270
16 19 70 304
Assuming W = 11, then L = 14 and P = 50 and A = 154 which is less than 180 which means that W can't be less than 12 because the area requirements would be violated.
Assuming W = 17, then L = 20 and P = 74 which is greater than 70 which means that W can't be greater than 16 because the perimeter requirements would be violated.
the answers to your questions are:
what are the smallest dimensions that the garden could have?
12 by 15
What would the largest that it could have?
16 by 19
you can actually graph these equations and visually see where the dimensions have to be.
we'll do it for your own information because some problems in the future will probably require graphing to solve them.
the perimeter equation becomes:
y = 4x + 6
the area equation becomes:
y = x^2 + 3x
we graph these 2 equation plus we graph:
y = 70
y = 180
the graph of these equations looks like this:
you can see from this graph that, in order for y to be >= 180, x has to be >= 12.
you can also see from this graph that, in order for y to be <= 70, x has to be <= 16.
the equation of the area is the curved line.
the equation of the perimeter is the straight line.
the y value for the curved line has to be greater than or equal to 180.
the y value for the straight line has to be less than or equal to 70.
the intersections of the horizontal lines with the equation lines tells you what the y value of the cutover point is.
the intersections of the vertical lines with the equation lines tells you what the x value of the cutover point is.
horizontal lines are easy to draw.
vertical lines are harder.
i can get close with the vertical lines, but not right on.
in this case i did pretty good.
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