SOLUTION: I have lawn beds 20 inches wide, by turning up the edges to form right angles. Find the depth that will maximize its cross-sectional area and allow the greatest amount of water to
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: I have lawn beds 20 inches wide, by turning up the edges to form right angles. Find the depth that will maximize its cross-sectional area and allow the greatest amount of water to
Log On
Question 105577: I have lawn beds 20 inches wide, by turning up the edges to form right angles. Find the depth that will maximize its cross-sectional area and allow the greatest amount of water to flow. What is the maximum cross-sectional area? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! I have lawn beds 20 inches wide, by turning up the edges to form right angles. Find the depth that will maximize its cross-sectional area and allow the greatest amount of water to flow. What is the maximum cross-sectional area?
:
Assume these things look this from the end:
|_|x
:
x = depth
(20-2x) = width
:
Area = x(20 - 2x)
:
A = -2x^2 + 20x; a quadratic equation, find the axis of symmetry and the vertex
:
The axis of symmetry equation: x = -b/(2a)
In our equation: a = -2 and b = +20
x =
x = -20/-4
x = +5 is the depth with for max water flow
:
Find the max cross sectional area; substitute 5 for x in the our equation
A = -2(5^2) + 20(5)
A = -2(25) + 100
A = -50 + 100
A = +50 sq inches is the max cross-sectional area (Vertex)
:
How about this; did it make sense to you?
