word problem please? algebra?
eastgate consignments wants to build a rectangular area in a conrner for children to play in while their parents shop. they have 30ft of low fencing. what is the maximum area they can enclose? what dimensions will yield this area? explain please?
Comments
A square with the sides being 7.5, the total area would be 56.25
In comparison, say you tried any rectangle:
7x8, Area = 56
6x9, Area = 54
5x10, Area = 50
4x11, Area = 44
...... so, you can see that when you get to the square, that will maximize your area.
A square will give you the largest area of a rectangular shape, so the side would be 30/4 = 7.5 ft.
A square has the max area, so
4x = 30
x = 7.5
to prove, area of a rectangle is
xy = A
perimeter is
P = 2(x+y)
P = 2x+2y
2x = P - 2y
2A = 2xy
2A = (P - 2y)y
2A = Py - 2y²
A = ½Py - y²
differentating
A' = ½P - 2y
setting equal to 0
0 = ½P - 2y
solving for y
y = P/4
in other words a square.
a square has the max area of any rectangle, so it would be
7.5 by 7.5, so the area would be 56.25 (i did that in my head in like 2 seconds, lol)
for you explaination, just say a square has the max area for any rectangle if you are given the perimeter
it it was just optimizing the four sides then i would look at the first answer. but if i am reading the problem as correctly, then since it is in a corner i assume that there are two preexisting sides, in which case the answer would be be 15.