What dimensions will maximize the area of this rectangle?

Question

A rectangle in the first quadrant is bound by the curve y = 15 - 2x. What dimensions will maximize the area of this rectangle?

It is bound by the FIRST Quadrant. The boundaries are the first quadrant x and y axis and the equation presented

Answer 1

the area = xy

take the derivative of this area function and set = 0

(xy)' = xy' + y = x(-2) + 15-2x = 15-4x =0

4x=15

x=15/4 =3 3/4 = 3.75 = y/2 = 7.5/2

y=15-2x=15-2(15/4)=15-15/2=15/2 = 7.5 = 7 1/2

the rectangle's area =xy= 15/2 times 15/4 = 225/8= 28 1/8 = 28.125

to check that as the maximum area try slightly more and slightly less than x=15/4 and y=15/2

try x=4 and y=15-8=7 4x7=28 < 28 1/8

try x=3 and y=15-6=9 3x9=27< 28 1/8

taking the derivative of a function, such as an area function, and setting it equal to zero

will lead to finding local extrema, such as a local maximum which is also a global maximum in this case

You might be tempted to believe the maximum area of a rectangle is a square

but with y=15-2x a square would mean x=y. substitute that and get

x=15-2x

3x = 15

x = 15/3 = 5 = y

5 by 5 square has area 25 < 28 1/8

the maximum area is the mean y by the mean x

graph the equation you get a triangle with height y=15 and base x=7.5

take half of those values, and get y=7.5 and x = 3.75

7.5 times 3.75 = 28 1/8, the maximum possible area