a farmer has 500ft of fencing with which to enclose a rectangular field. A straight riverbank will be used as part of the fencing on one side of the field. Prove that the are of the field is greatest when the rectangle is square.
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One side of the rectangle is a length of the riverbank, let's say of length x.
Then the length (y) of one of the perpendicular sides to the X sides would be (500 - x) / 2.
(For example, if x=100, then length y would be (500 - 100) / 2 = 200.
Area = x * y = x * (500 - x) / 2 = 250 * x - x^2 / 2
"To find the value of x that gives an area A maximum, we need to find the first derivative dA/dx (A is a function of x)." (Source noted below.)
A = 250 x - x^2 / 2
dA/dx = 250 - (1/2 * 2) x = 250 - x
"If A has a maximum value, it happens at x such that dA/dx = 0." (Source noted below.)
Set dA/dx = 0, giving 0 = 250 - x so x = 250
So, the maximum area occurs when the river side length is 250 feet and the dimensions of the rectangle are 250, 125, 250, 125. (Not a square!)
But that can't be right, can it? A square should be the answer, right?
Well, let's test that.
If we used a square, then the length of one side would be 500/3, with the riverside supplying the fourth 500/3 length.
Area of rectangle of 250 by 125 = 250 * 125 = 31,250
Area of square of 500/3 = 500/3 * 500/3 =~ 27,778.
Why would this be? The answer is that you are getting free use of the river as one wall, so, increasing that side up to a point increases the area.
How about a little bigger or a little smaller?
If x = 251, then Area = 251 * 124.5 = 31249.5
If x = 249, then Area = 249 * 125.5 = 31249.5
So this one is the largest:
Area of rectangle of 250 by 125 = 250 * 125 = 31,250
Then the length (y) of one of the perpendicular sides to the X sides would be (500 - x) / 2.
(For example, if x=100, then length y would be (500 - 100) / 2 = 200.
Area = x * y = x * (500 - x) / 2 = 250 * x - x^2 / 2
"To find the value of x that gives an area A maximum, we need to find the first derivative dA/dx (A is a function of x)." (Source noted below.)
A = 250 x - x^2 / 2
dA/dx = 250 - (1/2 * 2) x = 250 - x
"If A has a maximum value, it happens at x such that dA/dx = 0." (Source noted below.)
Set dA/dx = 0, giving 0 = 250 - x so x = 250
So, the maximum area occurs when the river side length is 250 feet and the dimensions of the rectangle are 250, 125, 250, 125. (Not a square!)
But that can't be right, can it? A square should be the answer, right?
Well, let's test that.
If we used a square, then the length of one side would be 500/3, with the riverside supplying the fourth 500/3 length.
Area of rectangle of 250 by 125 = 250 * 125 = 31,250
Area of square of 500/3 = 500/3 * 500/3 =~ 27,778.
Why would this be? The answer is that you are getting free use of the river as one wall, so, increasing that side up to a point increases the area.
How about a little bigger or a little smaller?
If x = 251, then Area = 251 * 124.5 = 31249.5
If x = 249, then Area = 249 * 125.5 = 31249.5
So this one is the largest:
Area of rectangle of 250 by 125 = 250 * 125 = 31,250