# Derivative of y = $\sqrt{16x^2+5x+15}$

You are building a new house on a cartesian plane whose units are measured in miles. Your house is to be located at the point $(2,0)$. Unfortunately, the existing gas line follows the curve $y= \sqrt{16x^2+5x+16}$. It costs 400 dollars per mile to install new pipe connecting your house to the existing line. What is the least amount of money you could pay to get hooked up to the system?

My try: find the derivative of the equation to find its critical points. And the compute the distance from the house to the critical point (local minimum).

However, I'm unable to compute the derivative. Kindly help with explanation.

• $15$ or $16$ ??? – Yves Daoust Aug 31 '18 at 12:44

You don't need to take the derivative of $\sqrt{16x^2+5x+15}$ to solve this problem.

If you hook up your house to the existing gas line at $(x,y) = (x,\sqrt{16x^2+5x+16})$, then the square of the length of the new pipe is $L^2 = (x-2)^2+(y-0)^2 = 17x^2+x+20$. At what value of $x$ is this minimized?

Once you've figured out the optimal value of $x$, you can compute the minimum length of the new pipe, and thus, the minimum cost to connect your house to the existing gas line.

• Ok. Instead of differentiating the equation of the curve y itself, I should differentiate the distance between the house and a point x,y on the curve and minimize it. Thank you. – Minu Jul 6 '14 at 22:13
• I solved using the methods mentioned in previous answers and the answer came out to be 1788.12. I tried inputting 1787, 1788, 1787.62, 1788.12, but none answer were accepted. Whats the exact correct solution? – user3807691 Aug 31 '18 at 12:12

The square $S(x)$ of the distance from the point $(x,\sqrt{16x^2+5x+15})$ on the curve is given by $$S(x)=(x-2)^2+16x^2+5x+15.$$ Minimize this.

You can use the derivative in the following manner. First refer to the graph for general idea:

$\hspace{4cm}$

The shortest distance from the point $(2,0)$ to the curve will be along the normal line to the curve. Let $(x_0,y_0)$ be the tangent point. The slope of the tangent line at $(x_0,y_0)$ is: $$y'(x_0)=\frac{32x_0+5}{2\sqrt{16x_0^2+5x_0+16}}.$$ The equation of the normal line at $(x_0,y_0)$ is: $$y=-\frac{2\sqrt{16x_0^2+5x_0+16}}{32x_0+5}x+b$$ The normal line must pass through the point $(2,0)$, so the full normal line is $$y=-\frac{2\sqrt{16x_0^2+5x_0+16}}{32x_0+5}x+\frac{4\sqrt{16x_0^2+5x_0+16}}{32x_0+5}.$$ Finally, the normal line intersects the curve at $(x_0,y_0)$, so: $$\sqrt{16x_0^2+5x_0+16}=-\frac{2\sqrt{16x_0^2+5x_0+16}}{32x_0+5}x_0+\frac{4\sqrt{16x_0^2+5x_0+16}}{32x_0+5}$$ Dividing by ${16x_0^2+5x_0+16}$, we get: $$32x_0+5=-2x_0+4 \Rightarrow x_0=-\frac1{34}.$$