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daijahamaker062816
26.09.2019 •
Mathematics
Alight source is located over the center of a circular table of diameter 4 feet. (see picture below) find the height h of the light source such that the illumination i at the perimeter of the table is maximum when
where s is the slant, is the angle at which the light strikes the table and k is a constant.
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Ответ:
Also, radius of the table is 4'/2=2', therefore, using Pythagoras theorem,s^2=h^2+2^2 (1), and consequently,sin(alpha)=h/s=h/sqrt(h^2+2^2)(2)
Substitute (1) and (2) in (0), we can writeI(h)=K*(h/sqrt(h^2+4))/(h^2+4)=Kh/(h^2+4)^(3/2)
To get a maximum value of I, we equate the derivative of I (wrt alpha) to 0, orI'(h)=0or, after a few algebraic manipulations, I'(h)=K/(h^2+4)^(3/2)-(3*h^2*K)/(h^2+4)^(5/2)=K*sqrt(h^2+4)(2h^2-4)/(h^2+4)^3We see that I'(h)=0 if 2h^2-4=0, giving h=sqrt(4/2)=sqrt(2) feet above the table.
We know that I(h) is a minimum if h=0 (flat on the table) or h=infinity (very, very far away), so instinctively h=sqrt(2) must be a maximum.Mathematically, we can derive I'(h) to get I"(h) and check that I"(sqrt(2)) is negative (for a maximum). If you wish, you could go ahead and find that I"(h)=(sqrt(h^2+4)*(6*h^3-36*h))/(h^2+4)^4, and find that the numerator equals -83.1K which is negative (denominator is always positive).
An alternative to showing that it is a maximum is to check the value of I(h) in the vicinity of h=sqrt(2), say I(sqrt(2) +/- 0.01)we findI(sqrt(2)-0.01)=0.0962218KI(sqrt(2)) =0.0962250K (maximum)I(sqrt(2)+0.01)=0.0962218KIt is not mathematically rigorous, but it is reassuring, without all the tedious work.
Ответ:
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