SOLUTION: 41. Anthony has 10 ft of framing and wants to use it to make the largest rectangular picture frame possible. Find the maximum area that can be enclosed by his frame. 42. A b

Question 177860: 41. Anthony has 10 ft of framing and wants to use it to make the largest
rectangular picture frame possible. Find the maximum area that can
be enclosed by his frame.

42. A ball is thrown upward from ground level. Its height h, in feet, above
the ground after t seconds is h = 48t - 16t2. Find the maximum height
of the ball.

43. Suppose a parabola has a vertex in Quadrant IV and a , 0 in the
equation y = ax2 + bx + c. How many real solutions will the equation
ax2 + bx + c = 0 have?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
41. Anthony has 10 ft of framing and wants to use it to make the largest
rectangular picture frame possible. Find the maximum area that can
be enclosed by his frame.
------
Perimeter = 2(L + W)
10 = 2(L + W)
L+W = 5
L = 5-W
------------
Area = LW
Area = (5-W)W
Area = 5W - W^2
------------------
This is a quadratic with a = -1, b=5
The maximum area occurs when W = -b/2a = -5(2*-1) = 5/2
---
If W + L = 5 and W = 5/2, then L = 5/2
---------
So the maximum area is: A = LW = (5/2)^2 = 25/4 sq ft
===============================================================
42. A ball is thrown upward from ground level. Its height h, in feet, above
the ground after t seconds is h = 48t - 16t2. Find the maximum height
of the ball.
---
h(t) = 48t - 16t^2
Max occurs when t = -48/(2*-16) = 48/32 = 3/2 seconds
--
Max height = h(3/2) = 48(3/2)-16(3/2)^2 = 72 - 36 = 36 ft.
===============================================================
43. Suppose a parabola has a vertex in Quadrant IV and a , 0 in the
equation y = ax2 + bx + c. How many real solutions will the equation
ax2 + bx + c = 0 have?
Comment: Somthing is missing in your statement.
====================================================
Cheers,
Stan H.

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