# Problem of the Week #90 - December 16th, 2013

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#### Chris L T521

##### Well-known member
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three times as strong as the other, are placed 10 ft apart, where should an object be placed on the line between the sources so as to receive the least illumination?

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#### Chris L T521

##### Well-known member
Staff member
This week's question was answered correctly by Ackbach, anemone, MarkFL and marthbalarka.

You can find Ackbach's solution here (using standard calculus):

Let $L$ be the illumination (luminosity), $I$ be the intensity (strength), $r$ the distance from the source to the point in question, and $k$ the proportionality constant. Then our general equation is
$$L= \frac{kI}{r^{2}}.$$
We assume the total luminosity adds linearly, so that $L_{ \text{tot}}=L_{1}+L_{2}$. Let $x$ be the distance along the line between the sources as measured from the stronger source. Then we seek to minimize the quantity
$$L_{ \text{tot}}= \frac{3kI}{x^{2}}+ \frac{kI}{(10-x)^{2}}.$$
The minimization of this quantity should occur at the same location as the minimization of
$$y:= \frac{L_{ \text{tot}}}{kI}= \frac{3}{x^{2}}+ \frac{1}{(10-x)^{2}}.$$
The bounds on $x$ are $x \in (0,10)$, since the luminosity technically blows up at each source. The minimum will obviously not occur at either source. Setting $y'=0$ is the same as setting
$$-6x^{-3}+2(10-x)^{-3}= \frac{2x^{3}-6(10-x)^{3}}{x^{3}(10-x)^{3}}=0,$$
which only happens when
$$2x^{3}=6(10-x)^{3} \implies x= \sqrt{3}(10-x) \implies \boxed{x= \frac{10 \sqrt{3}}{1+\sqrt{3}} \, \text{feet} }.$$
Note that, since this value of $x$ is approximately $5.9$, it is in the interval. Also note that
$$y''=18x^{-4}+6(10-x)^{-4}>0;$$
it follows that we found a local minimum. Note also that the other two roots of the cubic are complex and therefore do not contribute critical points.

You can find Mark's solution here (using Lagrange Multipliers):

Let's work this problem in general terms, and obtain a formula into which we can then plug our given data. Let:

$$\displaystyle 0<I_1$$ be the illumination of first light source

$$\displaystyle 0<I_2$$ be the illumination of second light source

$$\displaystyle 0<D$$ be the distance between the light sources

$$\displaystyle x$$ be the object's distance from the first light source

$$\displaystyle y$$ be the object's distance from the second light source

Thus, our objective function, which is the total illumination received by the object is:

$$\displaystyle f(x,y)=I_1x^{-2}+I_2y^{-2}$$

subject to the constraint:

$$\displaystyle g(x,y)=x+y-D=0$$

Using Lagrange multipliers, we obtain the system:

$$\displaystyle -2I_1x^{-3}=\lambda$$

$$\displaystyle -2I_2y^{-3}=\lambda$$

From this we find:

$$\displaystyle \lambda=-2I_1x^{-3}=-2I_2y^{-3}\implies y=\sqrt{\frac{I_2}{I_1}}x$$

Substituting for $y$ into the constraint, we obtain:

$$\displaystyle x+\sqrt{\frac{I_2}{I_1}}x=D$$

$$\displaystyle x=\frac{D}{\sqrt{\frac{I_2}{I_1}}+1}$$

Hence:

$$\displaystyle y=\sqrt{\frac{I_2}{I_1}}\left(\frac{D}{\sqrt{\frac{I_2}{I_1}}+1} \right)=\frac{D}{\sqrt{\frac{I_1}{I_2}}+1}$$

Since the objective function is unbounded at the boundaries $(x,y)=(0,D),\,(D,0)$, we are assured these values for $x$ and $y$ are at the minimum.

Using the given data (where we define the second light source to be the stronger):

$$\displaystyle I_2=3I_1$$

$$\displaystyle D=10\text{ ft}$$

we find:

$$\displaystyle x=\frac{10}{\sqrt{3}+1}\text{ ft}\approx4.09458563186124\text{ ft}$$

$$\displaystyle y=\frac{10}{\sqrt{\frac{1}{3}}+1}\text{ ft}\approx5.90541436813876\text{ ft}$$

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