Answer:
Step-by-step explanation:
Given:
Point Q = Q is on the north bank 1200 m east of P $40/m to lay cable underground and $80/m way to lay the cable.
Lets denote T be a point on north bank, therefore, point T is t m east and 100m north of P.
Note that 0 < t < 1200
Distance from P to T using Pythagoras theorem:
PT^2 = PQ^2 + QT^2
PT = √(t²+100²) = √(t²+10,000)
Distance from T to Q:
TQ = 1200 - t
From above, Cable from P to T is underwater, and costs $80/m to lay.
Cable from T to Q is underground, and costs $40/m to lay.
Total cost of the cable:
C(t) = 80√(t²+10,000) + 40(1200 - t)
Cost is at its minimum: when dC/dt = C'(t) = 0 and d^2C/dt^2 = C''(t) > 0
dC/dt = C'(t)
= 80 * 1/2 (t² + 10,000)^(-1/2) × 2t + 40 × (-1)
dC/dt = C'(t)
= 80t/√(t² + 10,000) - 40 = 0
80t/√(t² + 10,000) = 40
80t = 40√(t² + 10,000)
2t = √(t² + 10,000)
square both sides:
4t² = t² + 10,000
3t² = 10,000
t² = 10,000/3
t = 100/√3
≈ 57.735
d^2C/dt^2 =
C''(t) = [80√(t² + 10,000) - 80t × t/√(t²+10,000)] /√(t²+10,000)
C''(t) = [(80(t²+10,000) - 80t²) / √(t²+10,000)] /√(t²+10,000)
C''(t) = 800,000 / (t² + 10,000)^(3/2)
C''(t) > 0 for all t
Therefore C(t) is minimized at t = 100/√3
C(100/√3) = 80√(10,000/3 + 10,000) + 40 × (1200 - 100/√3)
= 80 √(40,000/3) + 48,000 - 4000/√3
= 16000/√3 + 48,000 - 4000/√3
= 48,000 + 12,000/√3
= 48,000 + 4,000√3
= 4000 (12 + √3)
≈ 54,928.20
Minimum cost is $54,928.20
This occurs when cable is laid underwater from point P to point T(57.735 m east and 100m north of P) and then underground from point T to point Q (1200 - 57.735)
= 1142.265 m
. Multiply both sides by x-10 to get rid of the fraction. -2(x-10) = 4. Distribute to get -2x+20=4 and -2x = -16. That gives us an x value of 8.