The value [tex]x\approx 3.8[/tex] 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:
[tex]\overrightarrow{PA}\,\bullet\,\overrightarrow{PB} = \|\overrightarrow{PA}\|\|\overrightarrow{PB}\|\cdot \cos \theta[/tex] (1)
Where:
- [tex]\|\overrightarrow{PA}\|[/tex], [tex]\|\overrightarrow{PB}\|[/tex] - Magnitudes of [tex]\overrightarrow{PA}[/tex] and [tex]\overrightarrow{PB}[/tex].
- [tex]\theta[/tex] - Internal angle, in sexagesimal degrees.
The magnitudes of each are respectively defined by line segment length formula: [tex]A(x,y) = (3,1)[/tex], [tex]B(x,y) = (5,3)[/tex], [tex]P(x, y) = (x, 0)[/tex]
[tex]\overrightarrow{PA} = \sqrt{(3-x)^{2}+1^{2}}[/tex]
[tex]\overrightarrow{PA} = \sqrt{10-6\cdot x +x^{2}}[/tex] (2)
[tex]\overrightarrow{PB} = \sqrt{(5-x)^{2}+3^{2}}[/tex]
[tex]\overrightarrow{PB} = \sqrt{34-10\cdot x +x^{2}}[/tex] (3)
By (1), (2) and (3) we have the following expression:
[tex](3-x)\cdot (5-x) +3 = \sqrt{10-6\cdot x + x^{2}}\cdot \sqrt{34-10\cdot x + x^{2}}[/tex]
[tex]15-8\cdot x +x^{2} = \sqrt{(10-6\cdot x +x^{2})\cdot (34-10\cdot x + x^{2})}\cdot \cos \theta[/tex]
[tex]\theta = \cos^{-1} \frac{15-8\cdot x +x^{2}}{\sqrt{(10-6\cdot x + x^{2})\cdot (34-10\cdot x +x^{2})}}[/tex] (4)
From geometry we know that sum of internal angles in triangles equals 180°, which means that angle APB must meet this condition:
[tex]0 < \angle APB < 180[/tex]
In addition, we know that cosine function is a bounded function between -1 and 1, where [tex]\theta = 0^{\circ} \to 1[/tex], [tex]\theta = 90^{\circ}\to 0[/tex], [tex]\theta = 180^{\circ}\to -1[/tex]
A quick approach consists in graphing (4) against x. Outcome is described in the second image attached. By direct inspection, we see that [tex]x\approx 3.8[/tex] maximizes angle APB.
We kindly invite to check this question on optimization: https://brainly.com/question/4302495
