Question

Given points A(3,1) and B(5,3) find the point P in the interval [3 ,5] along the X-Axis that MAXIMIZES angle APB.

Answer 1

The value $x\approx 3.8$ maximizes angle APB.

In this question we need to determine the maximum possible angle APB, which can be determined by definition of dot product, that is to say:

$\overrightarrow{PA}\,\bullet\,\overrightarrow{PB} = \|\overrightarrow{PA}\|\|\overrightarrow{PB}\|\cdot \cos \theta$ (1)

Where:

• $\|\overrightarrow{PA}\|$, $\|\overrightarrow{PB}\|$ - Magnitudes of $\overrightarrow{PA}$ and $\overrightarrow{PB}$.
• $\theta$ - Internal angle, in sexagesimal degrees.

The magnitudes of each are respectively defined by line segment length formula: $A(x,y) = (3,1)$, $B(x,y) = (5,3)$, $P(x, y) = (x, 0)$

$\overrightarrow{PA} = \sqrt{(3-x)^{2}+1^{2}}$

$\overrightarrow{PA} = \sqrt{10-6\cdot x +x^{2}}$ (2)

$\overrightarrow{PB} = \sqrt{(5-x)^{2}+3^{2}}$

$\overrightarrow{PB} = \sqrt{34-10\cdot x +x^{2}}$ (3)

By (1), (2) and (3) we have the following expression:

$(3-x)\cdot (5-x) +3 = \sqrt{10-6\cdot x + x^{2}}\cdot \sqrt{34-10\cdot x + x^{2}}$

$15-8\cdot x +x^{2} = \sqrt{(10-6\cdot x +x^{2})\cdot (34-10\cdot x + x^{2})}\cdot \cos \theta$

$\theta = \cos^{-1} \frac{15-8\cdot x +x^{2}}{\sqrt{(10-6\cdot x + x^{2})\cdot (34-10\cdot x +x^{2})}}$ (4)

From geometry we know that sum of internal angles in triangles equals 180°, which means that angle APB must meet this condition:

$0 < \angle APB < 180$

In addition, we know that cosine function is a bounded function between -1 and 1, where $\theta = 0^{\circ} \to 1$, $\theta = 90^{\circ}\to 0$, $\theta = 180^{\circ}\to -1$

A quick approach consists in graphing (4) against x. Outcome is described in the second image attached. By direct inspection, we see that $x\approx 3.8$ maximizes angle APB.

We kindly invite to check this question on optimization: brainly.com/question/4302495