+31 Demonstrate that the largest rectangle that has a perimeter of units of P in a square.
+36915 Let
x= one side length
2x= length of TWO sides
P-2x = length of OTHER TWO side
(P-2x)/2 = length of ONE other side
Area = side 1 x side 2
= x * (P-2x)/2
= 1/2Px -2x^2/2
= 1/2 Px -x^2
This is an upside down 'U' shaped parabola.....we are asked to find the MAXIMUM value of this area equation
This will be where the slope = 0
First derivative defines slope SLOPE= 1/2 P - 2x Set this equal to ZERO
1/2 P - 2x = 0
2x=1/2 P
x = 1/4 P So the maximum area is acheived when x = 1/4 P
the SECOND side will be ( P- 2x)/2 = (P- 2(1/4P))/2 = 1/2P / 2 = 1/4P The same as the FIRST side
If all of the sides are equal (which we just shown) the rectangle with the biggest area and perimeter P is a SQUARE !
+36915 In case you have not yet learned about derivatives (to find the slope):
We can identify the minimum or maximum value of a parabola by identifying the y-coordinate of the vertex. You can use a graph to identify the vertex or you can find the minimum or maximum value algebraically by using the formula x = -b / 2a. This formula will give you the x-coordinate of the vertex.
the parabola equation we had was:
Area = 1/2 px - x^2
= -x^2 + 1/2 px a = -1 b = 1/2 p
the MAXIMUM of the parabolic area equation will be @ x = -b/2a = (-1/2p) / 2(-1) = 1/4 p and again we have the value of a side which gives maximum area is x = 1/4 p so the rectangle is a square....
