A rectangle has one corner in quadrant I on the graph of y4x

A rectangle has one corner in quadrant I on the graph of y=4-x^2, another at the orign, a third on the positive y-axis and the fourth on the positive x-axis Express the area A of the rectangle as a function of x. What is the domain of A?

Solution

the are of the rectangle is = xy (length*bredth)

but given y = 4 -x^2

so A(x) = x(4-x^2)

= 4x -4x^3

domain means all the values that \'x\' can take . here we have no restriction so

domain = (-infinite , infinite)

to maximize A(x)

find A\'(x)

A\'(x) = 4 - 4.3x^2

A(x) = 4 -12x^2

so finding min or max A\'(x) =0

4 - 12 x^2 =0

4 = 12 x^2

3x^2 =1

x^2 =1/3

x = +sqrt(1/3) or - sqrt(1/3)

so when x= -1/sqrt(3) the Area A(x) is maximum

A(-sqrt(1/3) = 4 (-sqrt(1/3) + 12 sqrt(1/3)

= 8sqrt(1/3)

= 8 / sqrt(3)