Answer:
i) Rs. 110
ii) Rs. 24200
Step-by-step explanation:
Let the number of times the ticket price goes up be x.
Then,
The quadratic equation formed will be
(70 + 10x)(300 - 20x)
= 21000 + 3000x - 1400x - 200x^2 = -200x^2 + 1600x +21000
Let y = -200x^2 + 1600x +21000
That is, -200x^2 + 1600x + (21000 - y) = 0 -------- (1)
a = -200, b = 1600, c = 21000 - y
x must be real since it is the number of times the ticket price goes up.
So, b^2 - 4ac >= 0
(1600)^2 - 4 * -200 * (21000 - y) >= 0
2560000 + 16800000 - 800y >= 0
-800y >= -19360000
y <= (-19360000)/(-800)
y <= 24200
Therefore, y is either less than or equal to 24200
Which means, y has maximum value at 24200.
Therefore maximum income generated will be 24200.
Now, y = 24200 in (1) gives,
-200x^2 + 1600x + (21000 - 24200) = 0
-200x^2 + 1600x - 3200 =0
-200 ( x^2 - 8x + 16) = 0
x^2 - 8x + 16 = 0
(x - 4)^2 = 0
That is, x = 4
i) Ticket price that will generate the most income for the Drama club
= 70 + 10x
= 70 + 10*4
= 70 + 40 = Rs.110
ii) Income expected to be brought in = Rs. 24200
We already obtained the maximum income above but you could verify it by substituting x = 4 in (1) and thereby, obtaining the maximum value of the quadratic equation.
I hope this helps. Please do let me know if there are errors in the procedure.