consider the graph of y=-x^2+2x+4 in the first quadrant/ an inscribed rectangle has one vertex on the origin, one on the x-axis, one on the y-axis, and one on the curve.

a)write an expression for the area of the rectangle.

b)find the dimensions of the rectangle that gives the maximum area.

help? please&thankyou. :]

Hopefully this is not too late for you.
Looking at the equation as it's posted there is no upper limit for the size of a rectangle.
There is an important but confusing clue.
the Word inscribed. A parabola is not a bounded curve but an area bounded by the axis and the curve we can make a rectangle.
A subtle change Y=-(x^2)+2x +4 or Y=(-1)x^2+2x+4 yields an inverted parabola in cartesian coordinates.

The area of a rectangle is height * width
or X*Y .. We know an equation for Y so X*Y =
Area = X*(-1 x^2 +2x+4) =-1 x^3+2x^2+4x

This cubic equation will have a maxima in the region of interest. ( the biggest area)

Find the first derivative
of this cubic equation =3x^2+4x+4
solve for X.. we find the maximum and minimums
the minimum is not interesting we find x=2

Check your work by plugging in a number slightly above and below 2 to see if x=2 is really the biggest until you trust equations
x=2 y= -(2^2)+2*2 + 4 =4

I hope this helps