find the point p on the line y = 6x that is closest to the point (74,0). what is the least distance between…

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Here, one point is $P=(x,6x)$ (since $y = 6x$) and the other is $(74,0)$. So $D=\sqrt{(x - 74)^2+(6x-0)^2}=\sqrt{x^{2}-148x + 5476+36x^{2}}=\sqrt{37x^{2}-148x + 5476}$.

Step2: Minimize the distance function

Since minimizing $D$ is equivalent to minimizing $D^{2}=37x^{2}-148x + 5476$. Take the derivative of $f(x)=37x^{2}-148x + 5476$ with respect to $x$. $f^\prime(x)=74x-148$. Set $f^\prime(x) = 0$, then $74x-148=0$, which gives $x = 2$. Substitute $x = 2$ into $y = 6x$ to get $y=12$, so the point $P=(2,12)$.

Step3: Calculate the least - distance

Use the distance formula with $P=(2,12)$ and $(74,0)$. $D=\sqrt{(74 - 2)^2+(0 - 12)^2}=\sqrt{72^{2}+(- 12)^{2}}=\sqrt{5184 + 144}=\sqrt{5328}\approx73.0$

Answer:

$73.0$