Question

You are building a new house on a cartesian plane whose units are measured in miles.

Answer 1

We first find the shortest distance between the point (2, 0) and the curve $y=\sqrt{16x^2+5x+16}$
Let the point on the curve for which the line making the shortest distance be (x, y), then
$d= \sqrt{(x-2)^2+(y-0)^2} = \sqrt{(x-2)^2+y^2}$
But, $y=\sqrt{16x^2+5x+16}$
Therefore,
$d=\sqrt{(x-2)^2+(\sqrt{16x^2+5x+16})^2}=\sqrt{x^2-4x+4+16x^2+5x+16}\\= \sqrt{17x^2+x+20}$
For minimum d(d)/dx = 0
$\frac{d(d)}{dx} = \frac{34x+1}{2 \sqrt{17x^2+x+20}} =0\\34x+1=0\\x=- \frac{1}{34}$
Therefore, shortest distance is
$d= \sqrt{17(- \frac{1}{34})^2- \frac{1}{34}+20 } = \sqrt{ \frac{1359}{68}} =4.47 \ miles$
Since it costs 400 dollars per mile, 4.47 miles will cost 4.47 x 400 = $1,788.20

Answer 2

Closest point on the curve y = √(16 x² + 5 x + 16 ) is: ( 0, 4 )
For x =0, y = 4.
Distance from a house:
d = 2² + 4² = 4 + 16 = 20
d = √20 = 4.47213 miles
Costs:  4.47213  x  $400 = $1,788.85
The least amount of money to be payed is $1,788.85.