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 $m$.

$\dfrac{(4-m)^3}6 = \dfrac{16}3$

$(4-m)^3 = 32$

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

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