Mathematics Question and Solution
Calculating x, y = 12.
y = (8x)/√(2x+4)
12√(2x+4) = 8x
3√(2x+4) = 2x
9(2x+4) = 4x²
18x+36 = 4x²
4x²-18x-36 = 0
2x²-9x-18 = 0
2x²-12x+3x-18 = 0
2x(x-6)+3(x-6) = 0
Therefore;
x ≠ -3/2 units.
x = 6 units.
Calculating gradient of the curve at the tangent at x = 6, observing quotient rule.
It implies;
Gradient at the tangent, m is;
m = 5/4
Calculating n, gradient at the normal.
mn = -1
And m = (5/4)
Therefore;
(5/4)n = -1
n = (-⅘)
It implies;
(12-0)/(3-a) = n
Calculating a, the maximum x value at the normal.
(12-0)/(6-a) = (-⅘)
60 = -24+4a
84 = 4a
a = 21 units.
b = 21-6
b = 15 units.
c² = 12²+15²
c² = 369
c = 3√(41) units.
d = ½(c)
d = 1.5√(41) units.
tane = 12/15
e = atan(12/15)°
sin(atan(12/15)) = f/1.5√(41)
f = 6 units.
Where f is the width of area Orange.
cos(atan(12/15)) = g/1.5√(41)
g = 7.5 units.
Calculating x when y = 6.
6√(2x+4) = 8x
3√(2x+4) = 4x
9(2x+4) = 16x²
16x²-18x-36 = 0
8x²-9x-18 = 0
(x-(9/16))² = (18/8)+(81/256)
(x-(9/16))² = (657/256)
x = (9/16)±(3√(73)/16)
x = (9+3√(73))/16 units.
h = 6-x
h = 6-(9+3√(73))/16
h = (96-9-3√(73))/16
h = (87-3√(73))/16 units.
j = ½(15)+h
j = ½(15)+(87-3√(73))/16
j = (120+87-3√(73))/16
j = (207-3√(73))/16 units.
Where j is the length of area Orange.
Therefore;
Area Orange exactly is;
6*(207-3√(73))/16
= ⅛(621-9√(73)) square units.
= 68.0129957865 square units exactly in decimal.
