Arectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. 568 feet of fencing is used. find the dimensions of the playground that maximize the total enclosed area. remember to reduce any fractions and simplify your answers as much as possible.

First, we write an equation to represent that the fencing lengths add up to 568 feet. we call the side of the fence that has three segments of its length x and the side with only two segments y. We write 3x + 2y = 568. We also know that the area of the rectangle is equal to xy, so area = xy. We put y in terms of x using our first equation and find that y = (568 - 3x)/2. We plug this into our area equation and find that area = (568x - 3x^2)/2. To find the maximum we set the derivative equal to 0 and end up with 0 = 284 - 3x. We solve for x and get 94 and 2/3. We then put that into our first equation to find y = 142. So the dimensions that maximize the area are 94 2/3 x 142.

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

0 = ( 2 - 4 ) ( p + q )0 = ( 2 - 4 ) ( 2 + 5 i ),   by multiplying these0 = ( 4 + 10 i - 8 - 20 i )0 = 4 - 8 + 10 i - 20 i0 = - 4 -10 i   0 = 2 ( - 2 - 5 i )0 / 2 = - 2 - 5 i- 2 - 5 i = 0- 5 i = 2i = 2 / - 5