MI PU2 Promo 9758/2019/02/Q1 Town A, Town B and Town C are located in Wakandi Country. The distance between Town A and Town B is 60 km and the distance between Town B and Town C is 20 km. A railroad connects Town A to Town B (see diagram). A manufacturer plans to deliver a certain number of containers of its goods daily from Town A to Town C. To support this plan, the Wakandi government decides to build Interchange Station S and a road connecting this station to Town C (see diagram). Once the road is built, the goods manufacturer can deliver its containers from Town A to Town C by a combination of rail and road via Interchange Station S. The cost to deliver the containers daily by rail is $200 per km and the cost to deliver the containers daily by road is $300 per km.

1. Find the road distance $SC$ in terms of $x$ using Pythagoras' theorem.

From the diagram, the distance from Interchange Station S to Town B is $SB = 60 - x$.
Since $\angle SBC = 90^\circ$, we use Pythagoras' theorem in right-angled $\triangle SBC$:
$SC^2 = SB^2 + BC^2$
$SC = \sqrt{(60 - x)^2 + 20^2}$
$SC = \sqrt{3600 - 120x + x^2 + 400}$
$SC = \sqrt{x^2 - 120x + 4000}$

2. Formulate the total delivery cost $T$ using the given rates for rail and road.

The total daily delivery cost $T$ is the sum of the cost by rail (for distance $AS$) and the cost by road (for distance $SC$).
$T = 200(AS) + 300(SC)$
Since $AS = x$:
$T = 200x + 300\sqrt{x^2 - 120x + 4000} \quad \text{(shown)}$

1. Find the first derivative $\frac{\mathrm{d}T}{\mathrm{d}x}$ using the chain rule.

$T = 200x + 300(x^2 - 120x + 4000)^{\frac{1}{2}}$
$\frac{\mathrm{d}T}{\mathrm{d}x} = 200 + 300 \left(\frac{1}{2}\right)(x^2 - 120x + 4000)^{-\frac{1}{2}}(2x - 120)$
$\frac{\mathrm{d}T}{\mathrm{d}x} = 200 + \frac{300(x - 60)}{\sqrt{x^2 - 120x + 4000}}$

2. Set the derivative to zero and rearrange to isolate the square root.

For a stationary value, $\frac{\mathrm{d}T}{\mathrm{d}x} = 0$:
$200 + \frac{300(x - 60)}{\sqrt{x^2 - 120x + 4000}} = 0$
$\frac{300(x - 60)}{\sqrt{x^2 - 120x + 4000}} = -200$
$3(x - 60) = -2\sqrt{x^2 - 120x + 4000}$

3. Square both sides to form a quadratic equation and solve for $x$.

$(3(x - 60))^2 = \left(-2\sqrt{x^2 - 120x + 4000}\right)^2$
$9(x^2 - 120x + 3600) = 4(x^2 - 120x + 4000)$
$9x^2 - 1080x + 32400 = 4x^2 - 480x + 16000$
$5x^2 - 600x + 16400 = 0$
$x^2 - 120x + 3280 = 0$
Using the quadratic formula:
$x = \frac{120 \pm \sqrt{(-120)^2 - 4(1)(3280)}}{2}$
$x = \frac{120 \pm \sqrt{14400 - 13120}}{2} = \frac{120 \pm \sqrt{1280}}{2}$
$x = 60 \pm 8\sqrt{5}$

4. Reject the extraneous root and state the valid value of $x$.

From the equation $3(x - 60) = -2\sqrt{x^2 - 120x + 4000}$, the right side is negative (since the principal square root is positive). This requires $x - 60 < 0$, so $x < 60$.
Alternatively, S lies between Town A and B, so $0 \le x \le 60$.
Therefore, we reject $x = 60 + 8\sqrt{5} \approx 77.89$.
$x = 60 - 8\sqrt{5} \approx 42.11 \text{ (to 2 d.p.)}$

5. Use the first derivative test to verify that the value of $x$ gives a minimum $T$.

Let $x \approx 42.11$. We test values slightly below and above this point:
When $x = 42$:
$\frac{\mathrm{d}T}{\mathrm{d}x} = 200 + \frac{300(42 - 60)}{\sqrt{42^2 - 120(42) + 4000}} \approx -0.69 < 0$
When $x = 43$:
$\frac{\mathrm{d}T}{\mathrm{d}x} = 200 + \frac{300(43 - 60)}{\sqrt{43^2 - 120(43) + 4000}} \approx 5.71 > 0$
Since $\frac{\mathrm{d}T}{\mathrm{d}x}$ changes sign from negative to positive as $x$ increases through the stationary point, $T$ is a minimum.