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 $D$ 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 $(x,y)$ on the line $y = 6x$ and the other is $(74,0)$. So $D=\sqrt{(x - 74)^2+(6x-0)^2}=\sqrt{x^{2}-148x + 5476+36x^{2}}=\sqrt{37x^{2}-148x + 5476}$. The domain of $x$ (interval of interest) for this problem is $(-\infty,\infty)$ since $x$ can take any real - valued number on the line $y = 6x$.

Step2: Minimize the function

To find the minimum of $D(x)=\sqrt{37x^{2}-148x + 5476}$, we can instead minimize the function $f(x)=37x^{2}-148x + 5476$ (since minimizing $f(x)$ is equivalent to minimizing $D(x)$ as $D(x)=\sqrt{f(x)}$ and the square - root function is a monotonic increasing function for non - negative values). First, find the derivative of $f(x)$ using the power rule. $f^\prime(x)=74x-148$.

Step3: Set derivative equal to zero

Set $f^\prime(x) = 0$. So $74x-148 = 0$. Solving for $x$ gives $x = 2$.

Step4: Find the $y$ - coordinate

Since $y = 6x$, when $x = 2$, $y=6\times2 = 12$. So the point $P$ is $(2,12)$.

Step5: Calculate the least distance

Substitute $x = 2$ into the distance formula $D=\sqrt{37x^{2}-148x + 5476}$. $D=\sqrt{37\times2^{2}-148\times2 + 5476}=\sqrt{37\times4-296 + 5476}=\sqrt{148-296 + 5476}=\sqrt{5328}=6\sqrt{148}=12\sqrt{37}$.

Answer:

The point $P$ is $(2,12)$ and the least distance is $12\sqrt{37}$. The interval of interest of the objective function is $(-\infty,\infty)$.