Amanda has 400 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plu

Answer 1

You're missing some information.
Let L = length and W = width, then the perimeter is
P = 2L + 2W
You are given that P = 400.
You have not actually asked a question.

Answer 2

lets call the width x. Then the length might want to be one hundred fifty-x, as width + length = 1/2 of total perimeter. So the realm is x(one hundred fifty-x) = 150x - x^2. Now, the subsequent step must be finished in dissimilar thoughts. in case you comprehend any calculus, you may differentiate to get one hundred fifty - 2x = 0, ie x = seventy 5. in spite of the undeniable fact that, from the sounds of your question you want a extra algebraic way. if so, lets try ending up the sq.: A = 150x - x^2 x^2 - 150x + A = 0 (x^2 - 150x + seventy 5^2) + A = seventy 5*seventy 5 (x-seventy 5)^2 + A = seventy 5^2 A = seventy 5^2 - (x-seventy 5)^2. So A is unquestionably decrease than seventy 5^2, so the optimal might want to take position at the same time as the 2d bit is 0, ie x = seventy 5. Then the realm is seventy 5^2.

Answer 3

and the distance between framing positions is 24 inches? so divide the 400 by the plus and you will be on your way