**Problem**

The line y = 2x + 8 intersects the parabola y = x^{2} at points A and B. Point C is on the parabolic arc AOB where O is the origin. Locate C to maximize the area of the triangle ABC.

A. (1.1, 1.21)

B. (1, 1)

C. (0.9, 0.81)

D. (1.2, 1.44)

**Solution**

$y = y$

$x^2 = 2x + 8$

$x^2 - 2x - 8 = 0$

$x = 4 \text{ and } -2$

For x = 4

$y = 4^2 = 16$

The point is (4, 16)

For x = -2

$y = (-2)^2 = 4$

The point is (-2, 4)

Using the formula for the area polygon by coordinates

$A = \dfrac{1}{2}\left| \begin{matrix} x_1 & x_2 & x_3 & \cdots & x_n & x_1 \\ y_1 & y_2 & y_3 & \cdots & y_n & y_1 \end{matrix} \right|$

Rotate counterclockwise from point A

$A = \dfrac{1}{2}\left| \begin{matrix} x_A & x_C & x_B & x_A \\ y_A & y_C & y_B & y_A \end{matrix} \right|$

$A = \dfrac{1}{2}\left| \begin{matrix} -2 & x & 4 & -2 \\ 4 & y & 16 & 4 \end{matrix} \right|$

$A = \frac{1}{2} \left[ (-2y + 16x + 16) - (4x + 4y - 32) \right]$

$A = \frac{1}{2}(12x - 6y + 48)$

$A = 6x - 3y + 24$

Substitute y = x^{2} of the parabola

$A = 6x - 3x^2 + 24$

Differentiate then equate to zero to maximize the area

$\dfrac{dA}{dx} = 6 - 6x = 0$

$x = 1$

When x = 1

$y = 1^2 = 1$

Thus, point C is at (1, 1). *answer: B*

**Another Solution**

$A = \frac{1}{2}bh$

The base b is fixed, hence, for largest area of ABC, h must be longest. Altitude h is longest if C is farthest from line AB. Point C is farthest from line AB if the slope of the tangent line at point C is equal to the slope of the line AB.

From the given line

$y = 2x + 8$

$y' = 2$

From the parabola

$y = x^2$

$y' = 2x$

Substitute y' = 2

$2 = 2x$

$x = 1$

When x = 1

$y = 1^2 = 1$

Thus, point C is at (1, 1). *answer: B*