Given the circumference of a circle having equation $$ x^2+y^2-4x-8y+16=0 $$ consider a straight line that passes through the origin and intersects the circumference at the points $P$ and $Q$. Find the slope so that the area of the $APQ$ triangle is maximum. Point $A$ is $(0;4)$.
I know that I should use the derivative in order to determine the maximum area of the triangle, but I don't know how to approach the problem.
On
Define you straight line with slope m as y=m*x. Look for the intersection of this straight line with your circle. For a circle, this will obviously result in a quadratic equation:
$$ x^{2} + m^{2}x^{2} + 16 - 4x -8mx = 0 $$
The discriminant is $64m-48$. Consequently
$$ x_{1,2} = \frac{4+8m \pm \sqrt{64m-48}}{2(1+m^{2})} \\ y_{1,2} = mx_{1,2} \\ x_{3} = 0 \quad y_{3} =4 $$
The area of a simple polygon based on its coordinates can most easily be found by using the shoelace formula: $$ \frac{1}{2}∣x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}−x_{1}y_{3}−x_{3}y_{2}−x_{2}y_{1}∣ $$
this will be a formula in function of m. Differentiate and set equal to zero.
Good luck!
*fixed typo
On
Let $y=mx$, then the perpendicular distance of $A$ from $PQ$ (i.e. $y=mx$) is
$$h=\frac{4}{\sqrt{1+m^2}}$$
Now
\begin{align} 0 &= (1+m^2)x^2-4(2m+1)x+16 \\ (x_1-x_2)^2 &= (x_1+x_2)^2-4x_2x_2 \\ &= \frac{16(2m+1)^2}{(1+m^2)^2}-\frac{4 \times 16}{1+m^2} \\ &= \frac{16(4m-3)}{(1+m^2)^2} \\ S &= \frac{h\times PQ}{2} \tag{area} \\ &= 2\sqrt{\frac{(x_1-x_2)^2+(y_1-y_2)^2}{1+m^2}} \\ &= 2\sqrt{\frac{(x_1-x_2)^2+m^2(x_1-x_2)^2}{1+m^2}} \\ &= 2|x_1-x_2| \\ &= \frac{8\sqrt{4m-3}}{1+m^2} \\ \frac{dS}{dm} &= \frac{16}{(1+m^2)\sqrt{4m-3}}-\frac{16m\sqrt{4m-3}}{(1+m^2)^2} \\ 0 &= \frac{16(1+3m-3m^2)}{(1+m^2)^2\sqrt{4m-3}} \\ m &= \frac{3+\sqrt{21}}{6} \tag{$m > \frac{4}{3}$} \\ \end{align}
Let $m$ be the slope of the line through the origin. Find the coordinates of $P$ and $Q$ as a function of $m$, then find the area of the triangle as a function of $m$. Differentiate, set to zero ....