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Exercise 144Calculus › Single-variable calculus › Differentiation › Optimisation and curve sketchingL2Added 4 Nov 2026

At which height should a lamp be hung?

#optimisation#physics#illumination#Lambert's law

Statement

A lamp hangs at height hh above a table. At a point of the table at horizontal distance aa from the foot of the lamp, the illumination is I=kcos⁡θd2I=\dfrac{k\cos\theta}{d^2}, where dd is the distance to the lamp and θ\theta the angle between the light ray and the vertical. At which height is the illumination at this point maximal?

Hint

Express dd and cos⁡θ\cos\theta using aa and hh: d2=a2+h2d^2=a^2+h^2 and cos⁡θ=hd\cos\theta=\frac hd.

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Solution

Illumination as a function of hh.
I(h)=k h(a2+h2)3/2,h>0.I(h)=\frac{k\,h}{(a^2+h^2)^{3/2}},\qquad h>0.

Derivative.
I′(h)=k (a2+h2)−3h2(a2+h2)5/2=k a2−2h2(a2+h2)5/2.I'(h)=k\,\frac{(a^2+h^2)-3h^2}{(a^2+h^2)^{5/2}}=k\,\frac{a^2-2h^2}{(a^2+h^2)^{5/2}}.
Positive for h<a2h<\frac a{\sqrt2} and negative afterwards.

Answer. The best height is h=a2≈0.71 ah=\dfrac a{\sqrt2}\approx0.71\,a, and
Imax⁡=2k33 a2.I_{\max}=\frac{2k}{3\sqrt3\,a^2}.

Remark. Two effects compete: a high lamp is far (the 1d2\frac1{d^2} factor), a low lamp shines at a grazing angle (the cos⁡θ\cos\theta factor). The compromise is not at all obvious without calculus.