# Math Help - Calculus - Lightbulbs

1. ## Calculus - Lightbulbs

The question is on the picture (:

2. Hello, Fnus!

We are told: . $I \:=\:\frac{kP}{d^2}$

Let $P$ be a point between the two sources.
. . Let $x$ = distance from "40".
. . Then $6-x$ = distance from "5".

The total intensity is: . $I \;=\;\frac{40k}{x^2} + \frac{5k}{(6-x)^2} \;=\;40kx^{-2} + 5k(6-x)^{-2}$

Then we have: . $I'\;=\;\frac{-80k}{x^3} + \frac{10k}{(6-x)^3} \;=\;0$

Multiply by $\frac{x^3(6-x)^3}{10k}\!:\;\;-8(6-x)^3 + x^3 \;=\;0$

This simplifies to: . $x^3-16x^2+96x-192 \;=\;0$

. . which factors: . $(x-4)(x^2-12x+48) \;=\;0$

. . and has the real root: . $\boxed{x \:=\:4}$

Since $I'' \;=\;\frac{240k}{x^4} + \frac{30k}{(6-x)^4}$ is always positive,
. . the function is always concave up.

Therefore, $x = 4$ provides minimum $I.$