<span>So the exact dimensions are
width = 13*sqrt(2)
height = 13*sqrt(2)/2
Approximate dimensions are
width = 18.38477631085024
height = 9.19238815542512
I am assuming that there's a formatting issue with this problem and the actual radius is 13, not 1313. With that in mind, I will work out the problem with a radius of 13 and simply provide an answer without explanation for a radius of 1313.
First, let's define the height of the rectangle as a function of its width. If we consider just half the width, we can create a right triangle where 1 leg is half the width, the other leg is the height, and the hypotenuse is the radius of the semicircle. So
h = sqrt(13^2 - (w/2)^2)
h = sqrt(169 - w^2/4)
The area of the rectangle will be
A = wh
Substituting the equation for h, we get
A = wh
A = w * sqrt(169 - w^2/4)
Since we're looking for a maximum, that will only happen when the slope of the area function is 0 which you can determine by the first derivative. So let's calculate the first derivative.
A = w * sqrt(169 - w^2/4)
A' = d/dw [ w * sqrt(169 - w^2/4) ]
A' = d/dw [ w ] * sqrt(169 - w^2/4) + w * d/dw [ sqrt(169-w^2/4) ]
A' = 1 * sqrt(169 - w^2/4) + 1/2(169 - w^2/4)^(0.5 - 1) * d/dw [ 169-w^2/4 ] * w
A' = sqrt(169 - w^2/4) + (-1/4 * d/dw [ x^2 ] + d/dw [ 169 ]) * w / (2*sqrt(169-w^2/4))
A' = sqrt(169 - w^2/4) + (-w/2)*w / (2*sqrt(169 - w^2/4))
A' = sqrt(169 - w^2/4) + -w^2/2 / (2*sqrt(169 - w^2/4))
A' = sqrt(169 - w^2/4) - w^2/(4*sqrt(169 - w^2/4))
A' = sqrt(169 - w^2/4)*(4*sqrt(169 - w^2/4))/(4*sqrt(169 - w^2/4)) - w^2/(4*sqrt(169 - w^2/4))
A' = 4(169 - w^2/4)/(4*sqrt(169 - w^2/4)) - w^2/(4*sqrt(169 - w^2/4))
A' = (676 - w^2)/(4*sqrt(169 - w^2/4)) - w^2/(4*sqrt(169 - w^2/4))
A' = ((676 - w^2) - w^2)/(4*sqrt(169 - w^2/4))
A' = (676 - 2w^2)/(4*sqrt(169 - w^2/4))
A' = (338 - w^2)/(2*sqrt(169 - w^2/4))
Now find the zeros of the derivative
0 = (338 - w^2)/(2*sqrt(169 - w^2/4))
0*(2*sqrt(169 - w^2/4)) = (338 - w^2)
0 = 338 - w^2
w^2 = 338
w = +/- 13*sqrt(2)
Now a negative width doesn't make sense for this problem, so the desired width is 13*sqrt(2) or approximately 18.38477631085024
Let's plug that value into the equation for the height:
h = sqrt(169 - w^2/4)
h = sqrt(169 - 13*sqrt(2)/4)
h = sqrt(169 - 338/4)
h = sqrt(169 - 169/2)
h = sqrt(169/2)
h = 13/sqrt(2)
h = 13*sqrt(2)/2
So the exact dimensions are
width = 13*sqrt(2)
height = 13*sqrt(2)/2
Approximate dimensions are
width = 18.38477631085024
height = 9.19238815542512
If the radius is truly 1313, then
width = exactly 1313*sqrt(2), approximately 1856.862407
height = exactly 1313*sqrt(2)/2, approximately 928.4312037</span>
You can't really determine the amount of computers from this question, they're obviously not going to have one computer for everyone so half third, or even a fourth of the school population
