Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
The equation of the curve is;
y = ½(3x²)+4
Point P is;
P(4, 28)
Calculating OQ (x)
y = ½(3x²)+4
dy/dx = 3x, and x = 4.
Therefore;
dy/dx (gradient) = 3 * 4
dy/dx = 12
Therefore;
(28-0)/(4-x) = 12
12x = 48-28
x = 20/12
x (OQ) = 5/3 units.
QXp = 4 - x = 4 - (5/3)
QXp = (12-5)/3
QXp = 7/3 units.
Area Orange is;
Area under the curve at points (0, 4) and (4, 28) - Area triangle with height 28 unit and base (7/3).
= 48 - ½(28)(7/3)
= 48 - ⅓(98)
= ⅓(144 - 98)
= ⅓(46) square units.
