Question

Let
`R` be the region bounded between the parabola
`y=4x-x^2` and the x-axis. Find
`m` so that the line
`y=mx` divides
`R` into two pieces of equal area.

Answer 1

First, we observe that

`4x-x^2 = x(4-x) = 0 \implies x=0 \text{ or } x = 4`

and

`4x-x^2 = 4-(x-2)^2 \le 4`

so that
`R` is in the first quadrant. Any line
`y=mx` that slices this region into two pieces must then have a slope between
`m=0` and
`m=4` (which is the slope of the tangent line to the curve through the origin).

The parabola and line meet at the origin, and again when

`4x - x^2 = mx \\\\ ~~~~ \implies x^2 + (m-4)x = x (x + m - 4) = 0 \\\\ ~~~~\implies x = 4-m`

with
`4x-x^2\ge mx` for
`0\le x\le4-m`.

Now, the total area of
`R` is

`\displaystyle \int_0^4 (4x-x^2) \, dx = \left(2x^2 - \frac{x^3}3\right)\bigg|_0^4 = \frac{32}3`

so that half the area is 16/3.

The area of the left piece (containing the origin) is

`\displaystyle \int_0^(4-m) ((4x-x^2) - mx) \, dx = \left(\frac{4-m}2 x^2- \frac{x^3}3\right)\bigg|_0^(4-m) = \frac{(4-m)^3}6`

Solve for
`m`.

`\frac{(4-m)^3}6 = \frac{16}3`

`(4-m)^3 = 32`

`4 - m = \sqrt[3]{32} = 2\sqrt[3]{4}`

`\boxed{m = 4 - 2\sqrt[3]{4} \approx 0.825}`