Seth has a piece of aluminum that is 10 in. wide and 12 ft. long. He plans to form a gutter with a rectangular cross section and an open top by folding up the sides. What dimensions of the gutter would maximize the amount of water that it can hold? Hint: Write a quadratic function for the cross-sectional area and find the vertex.
2007-03-08 08:59:35 · 4 answers · asked by sillyboys_trucksare4girls 2 in Science & Mathematics ➔ Mathematics
Okay, wow... we have some really "bright" people in here.
2007-03-08 09:02:54 · update #1
The area of the cross section is x(10-x)/2 where x is the bottom dimension and (10-x)/2 is the side dimension. Simplify and you get:
area = -x^2/2 + 5x
The first derivative is -x+5. The first derivative is zero when x = 5, which is the maximum point of the parabola. So the gutter to maximize cross section is 5" wide and 2.5" deep.
2007-03-08 09:09:12 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Are you 11 years previous? Your Maths instructor will possibly instruct you Quadratic equations interior the subsequent 2-3 years. Quad ability 4 and of their ordinary case, those equations have 4 words occasion 3x^2 + 5x - 7 = 0........(3x^2 is asserted 3x squared) that's a undeniable sort of an unknown (letter) squared, plus or minus varied the comparable unknown (letter), plus or minus a selection equals 0. then you definately'll a thank you to remedy them the two via using brackets and a formulation. solid success - i'm so happy you appreciate Maths. everytime you want to be taught something new, ask somebody who'd decide directly to help you.
2016-12-18 18:15:11 · answer #2 · answered by ? 4 · 0⤊ 0⤋
do people really name their kids seth?
2007-03-08 09:01:31 · answer #3 · answered by OwNaGeR 3 · 0⤊ 0⤋
i am not sure
2007-03-08 09:01:52 · answer #4 · answered by Anonymous · 0⤊ 0⤋