Question

Find the dimensions of the rectangle with the largest area if the base must be on the x-axis and
its other two corners are on the graph of:
(a) y=16-x²,-4<=x<=4
(b)x²+y²=1
(c)|x|+|y|=1
(d)y=cos(x),-pi/2<=x<=pi/2

Answer 1

a) let the top right vertex be (x,y)

then the area = 2xy
A = 2xy = 2x(16-x^2)
= 32x - 2x^3
dA/dx = 32 - 6x^2
= 0 for a max of A
6x^2 = 32
3x^2 = 16
x^2 = 16/3
x = 4/√3 , then y = 16 - 16/3 = 32/3

so the rectangle has a base of 2(4/√3) = 8/√3
and a height of 32/3

b) follow the same method
c) nice symmetry
http://www.wolframalpha.com/input/?i=%7Cx%7C%2B%7Cy%7C%3D1

let (x, y) be the point of contact
the straight line in first quadrant : y = -x + 1
follow the same steps as a)

d) y = cosx, again, nice symmetry,
make a sketch
Same kind of problem
A = 2xy
= 2x(cosx)
dA/dx = x(-sinx) + cosx
= 0 for a max of A
xsinx = cosx
hard to solve, I will use Wolfram
http://www.wolframalpha.com/input/?i=xsinx+%3D+cosx
x = .8603336 , y = .652185

the rectangle is 1.72067 by .652185