Probem49+Danny+G.+Sylvia

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 feet apart, where should an object be placed on the line between the sources so as to recieve the __least__ illumination. (minimize the function)

We know that AB = 10 ft and that Illumination = Strength / (distance)^2 We can choose a point C that is "y" feet from B, and (10-y) feet from A. Also, we can choose A to have 3x strength and point B to have x strength. But strength "x" is a constant.

Therefore, the illumination at point C is equal to both the illumination from point A and the illumination from point B

I = A + B I = (3x)/(10-y)^2 + x/y^2 I agree -- and remember that x is a constant here!

We must find the derivative of the equation and set it equal to zero in order to find critical value and possible extrema

I' = (10-y)^2(0) - (3(2(10-y)(-1))/(10-y)^4 + y^2(0) - (2y)/y^4 You are using the quotient rule for two different functions.

I' = 6x /(10-y)^3 - 2x /y^3 (and keep in mind again that x is a constant, so perhaps you can just factor it out!) 0 = 6/(10-y)^3 - 2/y^3 Now multiply to get rid of the fractions -- your answer will be in terms of a cube root but don't be afraid of it...or just use your calculator to find the value of y!Also, don't forget to see what I would be at the extremities (endpints of the domain!)

simplify 6/(10-y)^3 = 2/y^3 6y^3 = 2(10-y)^3 3y^3 = -y^3 + 30y^2 - 300y + 1000 0 = -4y^3 + 30y^2 -300y + 1000

Plug into calculator (substitute x for y) x = 0 when y = 4.0945856 There is a local max at y = 4.0945856

Minimize over [0,10] I(0) = 3x/ (10-0)^2 + x/ (0)^2 I(0) = 3x / 100 + 0 I(0) = (3/100)x I(0) = .03x

I(10) = 3x/ (10-10)^2 + x/ (10)^2 I(10) = 0 + (x/100) I(10) = (1/100)x I(10) = .01x

I(4.09) = 3x/(10-4.09)^2 + x/(4.09)^2 I(4.09) = 3x/(34.92) + x/(16.73) I(4.09) = .086x + .06x I(4.09) = .146x

Easier (above) : if you look at line 2, why not divide both sides by 2, and then take the cube root of both sides? That way you get rid of both y^3's! And once you have this value of y, you can plug it into I to see which gives you the lowest value! or largest? are you trying to maximize or minimize the illumination? -Ms. Sweeney, I have struggled finding the value of "y" that enables the function to equal zero.