Probem24+Mimi,+Ari

Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola y = 8 - x^2.

Let's say the x-coordinates are -x and x and the y-coordinate is 8 - x^2. length of the base = 2x height = 8 - x^2 the area of the rectangle is A = 2x(8 - x^2) = 16x - 2x^3 dA/dx = 16 - 6x^2 to get maximum area, dA/dx = 0 16 - 6x^2 = 0, so x = 1.63 when plugging in to the original equation for height, y = 8 - (1.63)^2 = 5.343 so the width would be 1.63 x 1.63 = 2.66 and the height would be 5.244. the area would be w x h = 2.66 x 5.244 = 13.95 (approximately)

We now have to take into consideration the end points for the domain of x. To do this we must find the x intercepts by setting the 8 - x^2 equal to 0: The smallest the base can be is 0, therefore the domain for x is (0, 2.828) Now lets plug in the end points in the the equation A =16x-2x^3, shall we!
 * 0 = 8 - x^2
 * x^2 = 8
 * x = __+__ 2sqrt(2), but we cant have a negative distance so x= 2sqrt(2)=2.828
 * A = 16(0) - 0 = 0
 * A = 16(2.828) - 2(2.828)^3 = 0.0137

So now we have our three contestents!


 * CONTESTANT A:** 13.95
 * CONTESTANT B:** 0
 * CONTESTANT C:** 0.0137 (practically zero if it wasnt for rounding)

Who's the biggest???????????????


 * CONTESTANT A!!!!!!!!!!**

The question asks for the dimensions so the answer is 2.66 x 5.244. 