Question

What point on the line y 9x+4 is closest to the origin? Let D be the distance between the two points. What is the objective function in terms of the x-coordinate? (Type an expression.)

Answer 1

Answer:

The objective function in terms of the x-coordinate is $d=\sqrt{82x^2+72x+16}$.

The point closest to the origin is $(-\frac{18}{41},\frac{2\sqrt{82}}{41})$.

Step-by-step explanation:

The formula for the distance from point (x, y) to the origin is

$d=\sqrt{x^{2}+y^{2} }$

So, in our case, $y=9x+4$ and the distance is

$d=\sqrt{x^{2}+(9x+4)^{2}  }\\\\d=\sqrt{x^2+81x^2+72x+16} \\\\d=\sqrt{82x^2+72x+16}$

This is the objective function.

Next, we need to find the derivative of the function

$d=\sqrt{82x^2+72x+16}\\\\\frac{d}{dx}d= \frac{d}{dx}\sqrt{82x^2+72x+16}\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\\\f=\sqrt{u},\:\:u=82x^2+72x+16\\\\\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(82x^2+72x+16\right)\\\\\frac{1}{2\sqrt{u}}\left(164x+72\right)\\\\\mathrm{Substitute\:back}\:u=82x^2+72x+16\\\\\frac{1}{2\sqrt{82x^2+72x+16}}\left(164x+72\right)\\\\\frac{d}{dx}d=\frac{2\left(41x+18\right)}{\sqrt{82x^2+72x+16}}$

Now, we set the derivative function equal to zero to find the critical points

$\frac{2\left(41x+18\right)}{\sqrt{82x^2+72x+16}}=0$

$\frac{f\left(x\right)}{g\left(x\right)}=0\quad \Rightarrow \quad f\left(x\right)=0\\\\2\left(41x+18\right)=0\\\\\frac{2\left(41x+18\right)}{2}=\frac{0}{2}\\\\41x+18=0\\\\x=-\frac{18}{41}$

We need to check that the value that we found is a minimum point for this we analyze the intervals of increase or decrease (First derivative test).

We can use a sign chart. In a sign chart, we pick a test value at each interval that is bounded by the critical points and check the derivative's sign on that value.

This is the sign chart for our function:

$\left\begin{array}{ccc}\mathrm{Interval}&\mathrm{Test \:x-value}&f'(x)\\(-\infty,-\frac{18}{41} )&-1&-9.02\\(-\frac{18}{41},\infty )&1&9.05\end{array}\right$

d(x) decreases before $x=-\frac{18}{41}$, increases after it, and is defined at $x=-\frac{18}{41}$. So d(x) has a relative minimum point at $x=-\frac{18}{41}$.

The point closest to the origin is

$d=y=\sqrt{82(-\frac{18}{41})^2+72(-\frac{18}{41})+16}=\frac{2\sqrt{82}}{41}$

$(-\frac{18}{41},\frac{2\sqrt{82}}{41})$