Respuesta :
The minimum value of a function can be obtained from its derivative
The point on the x-axis that makes [tex]\overline{AC}[/tex] + [tex]\overline{BC}[/tex] minimum is the point [tex]\underline{(9, \, 0)}[/tex]
Reason:
Let (d, 0) represent the coordinate of the point, on the x-axis, we have;
[tex]\overline{AC}[/tex]² = (4 - d)² + (-5)²
[tex]\overline{AC}[/tex] = √((4 - d)² + (-5)²)
[tex]\overline{BC}[/tex]² = (12 - d)² + 3²
[tex]\overline{BC}[/tex] = √((12 - d)² + 3²)
[tex]\overline{AC}[/tex]² + [tex]\overline{BC}[/tex]² = L = 4² + (-5 - d)² + 12² + (3 - d)²
When [tex]\overline{AC}[/tex] + [tex]\overline{BC}[/tex] is minimum, we have;
[tex]\lim_{n \to \infty} a_n \dfrac{d}{dd} (\overline{AC} + \overline{BC}) = \dfrac{d}{dd} \left(\sqrt{ (4 - d)^2 + (-5)^2)} + \sqrt{ (12 - d)^2 + 3 ^2)} \right) = 0[/tex]
Which gives;
[tex]\dfrac{d}{dd} \left(\sqrt{ (4 - d)^2 + (-5)^2)} + \sqrt{ (12 - d)^2 + 3 ^2)} \right) = \dfrac{2 \cdot d - 24}{2 \cdot \sqrt{d^2-24 \cdot d + 153} } + \dfrac{2 \cdot d - 8}{2 \cdot \sqrt{d^2 - 8\cdot d + 41} }[/tex]
Therefore;
[tex]\dfrac{2 \cdot d - 24}{2 \cdot \sqrt{d^2-24 \cdot d + 153} } + \dfrac{2 \cdot d - 8}{2 \cdot \sqrt{d^2 - 8 \cdot d + 41} } = 0[/tex]
[tex]\dfrac{2 \cdot d - 24}{2 \cdot \sqrt{d^2-24 \cdot d + 153} } =- \dfrac{2 \cdot d - 8}{2 \cdot \sqrt{d^2 - 8 \cdot d + 41} }[/tex]
Squaring both sides and cross multiplying gives;
16·d² - 528·d + 3456 = 0
Which gives;
16·(d - 24)·(d - 9) = 0
Therefore, d = 24, and d = 9
The point (9, 0), is closer to the given points than the point (24, 0), therefore;
The point on the x-axis that makes [tex]\overline{AC}[/tex] + [tex]\overline{BC}[/tex] minimum is the point [tex]\underline{(9, \, 0)}[/tex]
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