On Reddit AskMath, a student posted about a question whose mark scheme did not seem correct. Redditors agreed there was a mistake in the exam. So let me present the question and explain where the mark scheme is wrong.

A ship is sailing with speed v kmh^-1. The sailing cost per hour, $C is given by:

C = v² + 3000/v + 100

(a) Find the speed that makes C a minimum. Justify the value of C is a minimum.

(b) Hence find the minimum sailing cost for a journey of 150 km.

As usual, watch the video for a solution. https://youtu.be/

Student Finds Error On A Cambridge Exam

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Answer To

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

A ship is sailing with speed v kmh^-1.

The sailing cost per hour, $C is given by:

C = v² + 3000/v + 100

(a) Find the speed that makes C a minimum. Justify the value of C is a minimum.

This part is straight-forward. Set the first derivative equal to 0 and solve.

C′(v) = 0
2v – 3000/v² = 0
v³ – 1500 = 0
v = ∛1500
(we ignore the complex roots since speed is a real value)

To justify this is a minimum, we can use the second derivative test.

C′′(v) = 2 + 6000/v³ > 0

The second derivative is always positive for positive speeds, so it will be positive for the speed v = ∛1500. Therefore this is a minimum.

(b) Hence find the minimum sailing cost for a journey of 150 km.

The “hence” in the question suggests to substitute the answer from part (a) into part (b). The cost per hour is:

C(∛1500)
= (∛1500)² + 3000/∛1500 + 100
≈ 493.111 $/h

But this is just a rate per hour. To find the cost we have to multiply by the time the journey takes.

t(v) = 150/v = 150/∛1500 ≈ 13.104 h

The total cost is the product of the time and the cost per hour:

t(v) · C(v)
(13.104 h)(493.111 $/h)
≈ $6462

This calculation matches the official mark scheme. But is this actually a minimum cost?

The problem with the answer

The minimum cost per hour is not going to be the minimum cost for the distance, as faster speeds may cost more per hour but actually take less time.

For a given speed v km per hour, a 150 km trip will take:

150/v hours

And it will cost:

C = v² + 3000/v + 100

So the total cost is the product of the two:

t(v) · C(v)
(150/v)(v² + 3000/v + 100)
150v + 450000/v² + 1500/v

Let F(v) be this function. We can minimize the cost by setting the first derivative equal to 0.

F′(v) = 0
150 – 9000000/v³ – 15000/v² = 0
v³ – 100v – 6000 = 0

Solving this cubic equation (using WolframAlpha) there is only 1 real root which is v = 20 km per hour.

This yields a minimum cost of:

F(20)
= 150(20) + 450000/(20)² + 1500/(20)
= $4875

This is the correct answer, and it is more than $1000 cheaper than the supposed minimum cost of the mark scheme.

Somewhere in creating the question the test-markers thought that the minimum cost per hour would be the minimum cost per distance.

We can verify a sailing speed of 20 km per hour does cost more per hour. We have C(20) = $625 per hour, which is a lot more than C(∛1500) = $493.

But going at the faster speed means a journey of only 150/20 = 7.5 hours, which is more than 5 hours saved time compared to 150/∛1500 = 13.104 hours.

Cambridge has yet to acknowledge the mistake in its mark scheme, and the student who posted the flaw was rebuffed by his teachers and school administration.

I have no idea how many students actually got the correct answer but were marked wrong because of a flawed mark scheme.

Please do share this video or post so we can raise awareness and get a correction to the mark scheme.

References

Reddit AskMath
https://www.reddit.com/r/askmath/comments/1txjvev/is_this_mark_scheme_wrong/

Cambridge IGCSE website
https://www.cambridgeinternational.org/programmes-and-qualifications/cambridge-upper-secondary/cambridge-igcse/

This puzzle has gotten popular because many people thought it was impossible to solve. Can you figure it out?

As usual, watch the video for a solution.

Inscribed Circle Puzzle

Or keep reading.

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Answer To Inscribed Circle Puzzle

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

The first step is to show the 4 right triangles are congruent. The angles of adjacent triangles are complementary, but so are angles in each of the right triangles, so all the acute angles in the right triangles are equal and the triangles are similar. As each hypotenuse is a side of the inscribed square, the hypotenuses are congruent. Similar triangles with a congruent side are congruent, so the 4 right triangles are congruent.

So all 4 triangles have an area of 24, and their sides can be labeled as a, b and c.

The inscribed’s square area is 100, so we have:

c² = 100
c = 10

In the large square, its area is equal to the square of its side length and the sum of the 5 shapes inside, so we have:

(a + b)² = 100 + 4(24)
(a + b)² = 196
a + b = 14

Finally we inscribe a circle in one of the triangles.

There are 2 formulas for the inradius we could use. For any triangle:

A = r(a + b + c)/2
24 = r(14 + 10)/2
r = 2

Or we can use that a right triangle’s inradius is given by:

c = a + b – 2r
10 = 14 – 2r
r = 2

In either case, the area of the inscribed circle is:

πr²
= π(2)²
= 4π

Deriving the inradius formulas

For any triangle:

For a right triangle:

Special thanks this month to:

Daniel Lewis
Lee Redden
Kyle

Thanks to all supporters on Patreon and YouTube!

Reference

Seen on Reddit AskMath
https://www.reddit.com/r/theydidthemath/comments/1r1yepq/can_you_solve_it_request/

As usual, watch the video for a solution.

How Tall Is The Man Puzzle

Or keep reading.

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Answer To How Tall Is The Man Puzzle

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

At first the puzzle seems impossible to solve as we do not know the distance between the two walls. Surprisingly, the distance does not matter! For poles of length a and b, the height of intersection h between the lines joining the top of one pole to the bottom of the other is equal to:

h = ab/(a + b)

We can set a = 4 and b = 6 to solve for the height of the man:

h = (4)(6)/(4 + 6) = 2.4 m

Proof of two poles formula

Label the two triangles are EFG and GJE, with the man replaced by the line segment KL. Let the distance between the two poles be d, and let KG = x so that KE = d – x, as shown below.

Triangles EFG and KLG are similar, giving:

x/h = d/a
x = dh/a

Triangles GJE and KLE are similar, giving:

(d – x)/h = d/b
bd – bx = dh
x = (bd – dh)/b

Both equations are equal to x, so setting them equal to each other gives:

dh/a = (bd – dh)/b
bdh = abd – adh
bdh + adh = abd
hd(b + a) = abd
h(b + a) = ab
h = ab/(a + b)

Special thanks this month to:

Kyle
Daniel Lewis
Lee Redden

Thanks to all supporters on Patreon and YouTube!

References

Puzzle
https://x.com/H0H0v/status/2007323697477210181

Two Poles Formula
https://artofproblemsolving.com/wiki/index.php/Two_poles_formula?srsltid=AfmBOootttF_6CfLUa7mr2vL_pqughPMQSXMEuLj0KEtGuWfJJ6E_JiR

Robert Wadlow
https://commons.wikimedia.org/wiki/File:Robert_Wadlow,_World%27s_Tallest_Man_Statue.jpg
KMOM14, CC BY 3.0 , via Wikimedia Commons