急 ! F4 Quadratic Functions

23.
(a)(i)
Length of the fence :
(1/4)(2πx) + y + (1/4)(2πx) = 100
(1/4)(2*3*x) + y + (1/4)(2*3*x) = 100
(3/2)x + y + (3/2)x = 100
y + 3x = 100
y = 100 - 3x ...... [1]

(a)(ii)
(1/4)π(x)² + xy + (1/4)π(x)² = A
(1/4)*3*x² + xy + (1/4)*3*x² = A
(3/4)x² + xy + (3/4)x² = A
A = (3/2)x² + xy ...... [2]

Put [1] into [2] :
A = (3/2)x² + x(100 - 3x)
A = (3/2)x² + 100x - 3x²
A = 100x - (3/2)x²

(b)
A = 100x - (3/2)x²
A = -(3/2)[x² - (100x)(2/3)]
A = -(3/2)[x² - (200x/3) + (100/3)²] + (3/2)(100/3)²
A = (5000/3) - (3/2)[x - (100/3)²]
For all real values of x, (3/2)[x - (100/3)]² ≥0
Hence, maximum A = 5000/3
Maximum area of the garden = 1667 m² (to the nearest integer)


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25.
(a)
q = 12000 - 50(p - 160)
q = 12000 - 50p + 8000
q = 20000 - 50p

(b)
Put $y be the income.

y = pq
y = p(20000 - 50p)
y = -50(p² - 400p)
y = -50(p² - 400p + 200²) + 50*200²
y = 2000000 - 50(p - 200)²
For all real value of p, 50(p - 200)² ≥ 0
Hence, maximum y when 50(p - 200)² = 0, i.e. p = 200
Price of ticket in order to get the maximum income = $200