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Two light sources of identical strength are placed $ 10 $ m apart. An object is to be placed at a point $ P $ on a line $ \ell $, parallel to the line joining the light sources and at a distance $ d $ meters from it (see the figure). We want to locate $ P $ on $ \ell $ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source.(a) Find an expression for the intensity $ I(x) $ at the point $ P $.(b) If $ d = 5 $ m, use graphs of $ I(x) $ and $ I'(x) $ to show that the intensity is minimized when $ x = 5 $ m, that is, when $ P $ is at the midpoint of $ \ell $.(c) If $ d = 10 $ m, show that the intensity (perhaps surprisingly) is not minimized at the midpoint.(d) Somewhere between $ d = 5 $ m and $ d = 10 $ m there is a transitional value of $ d $ at which the point of minimal illumination abruptly changes. Estimate this value of $ d $ by graphical methods. Then find the exact value of $ d $.

a) $I(x)=\frac{k}{x^{2}+d^{2}}+\frac{k}{(10-x)^{2}+d^{2}}=\frac{k}{x^{2}+d^{2}}+\frac{k}{x^{2}-20 x+100+d^{2}}$b) $I(x)$ has a minimum at $x=5 \mathrm{m}$c) $I(x)$ has is minimized at $x=0$ and $x=10 \mathrm{m} .$ And the midpoint is the most brightly lit point.d) 5$\sqrt{2}$ which is approximately 7.07

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

Missouri State University

Baylor University

University of Nottingham

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