The back of Tom's property is a creek. Tom would like to enclose a rectangular area, using the creek as one side and fencing
1 answer:
9514 1404 393
Answer:
1250 square feet
Step-by-step explanation:
If x is the length of the side perpendicular to the creek, then the third side is (100 -2x) = 2(50 -x). The area is the product of length and width:
A = x(2)(50-x)
We observe that this is a quadratic function with zeros at x=0 and x=50. The vertex (maximum) of a quadratic function is on the line of symmetry, halfway between the zeros. The value of x there is (0 +50)/2 = 25.
Then the maximum area is ...
A = (25)(2)(50 -25) = 1250 . . . . square feet
_____
<em>Additional comment</em>
Note that half the length of the fence is used in one direction (parallel to the creek), and half is used in the other direction (perpendicular to the creek). This 50/50 split is the generic solution to all sorts of rectangular corral problems, with or without a creek, with or without internal partitions.
Half the fence is perpendicular to the other half. (If the costs are different in different directions, then the cost is what is split 50/50.)
You might be interested in
To solve this problem you must apply the proccedure shown below:
1. You have the following equation of a parabola, given in the problem above:
x<span>=1/16y^2
2. Then, based on the graph attached, you have:
p=y^2/4x
p=8^2/(4)(4)
p=64/16
p=4
3. The directrix is:
directrix=h-p
directrix=0-4
directrix=-4
The answer is:-4</span>
1. You have the following equation of a parabola, given in the problem above:
x<span>=1/16y^2
2. Then, based on the graph attached, you have:
p=y^2/4x
p=8^2/(4)(4)
p=64/16
p=4
3. The directrix is:
directrix=h-p
directrix=0-4
directrix=-4
The answer is:-4</span>
The given quadratic describes a parabola that opens upward. Its one absolute extreme is a minimum that is found at x = -3/2. The value of the function there is
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at (-1.5, -3.25).
_____
For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at (-1.5, -3.25).
_____
For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.