what is the maximum area of a rectangle inside the parabola y=4-x^2 with its base on the x-axis.?

Answer 1

The parabola is opening downward. The space in question is bounded by the parabola and the X axis. If the rectangle has its base on the x-axis, then the upper corners are touching the parabola. Also, note that if we solve for X and find Y, then X * Y = ½ the rectangle (on the right side of the Y-axis. The other half is on the left side of the Y-axis. Therefore, the total area = 2X * Y.


The area of the rectangle = 2X * Y.
Y = 4-X^2
by substitution, then the rectangle
Area = 2X * -X^2

set the equation to 0

0 = X^2 - 2X

taking the derivative of the equation:

0 = 2X - 2

and solve for X;

X = 1

at X = 1, then Y = 4 - 1^2 = 3

The area under the parabola is

2X * Y = 2 * 3 = 6

Answer 2

2x = base of rectangle
y = 4 - x² = height of rectangle
A = area of rectangle

A = 2xy = 2x(4 - x²) = 8x - 2x³

Take the derivative and set equal to zero to find the critical values.

dA/dx = 8 - 6x² = 0
6x² = 8
x² = 8/6 = 4/3
x = 2/√3

The area of the rectangle is:

A = 8x - 2x³ = 8(2/√3) - 2(2/√3)³

A = 16/√3 - 16/(3/√3) = 32/(3/√3) ≈ 6.15840209

Answer 3

The area is 2*x*(4-x^2) = 8x-2x^3.
Now take derviative , and set to 0.
You get 8-6x^2=0, so x=sqr(8/6).

So, area is (8*sqr(8/6))-(2*(sqr(8/6))^3) =
((8sqr(8/6))-((16/6)sqr(8/6)) =
(32/6)*sqr(8/6) = 6.158 approximately.

But check caluculations!

Answer 4

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