A rectangle is constructed with its base on the diameter of a semicircle with radius 1313 and with its two other vertices on the semicircle. what are the dimensions of the rectangle with maximum area?

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