As part of a recent Network Meeting we looked at the question below which could be used with GCSE students (without calculus), or A level students (with calculus).
We looked at what assumptions students would need to make and what difficulties they may have, and briefly looked at the source of the question. Here are just two possible solutions:
Without calculus:
Let P be point (x, y) and greatest distance from origin be D .
Look for repeated root x2 + y2 = D2 and 3×2 + y2 + 15x = 18
Solve simultaneously: 2×2 + 15x + D2 – 18 = 0
Use b2 -4ac = 0
152 – (4)(2)(D2 – 18) = 0
225 + 144 = 8D2
D2 = 360/8
D = (Sqaure root equation)
With calculus:
D2 = x2 + y2 and 3×2 + y2 + 15x = 18
so D2 = 18 – 15x – 2×2
Let D2 = L
dL/dx = -15 – 4x with stationary point when dL/dx = 0, x = – 15/4
d2L/dx2 = -4 <0 so maximum at x = -15/4
When x = -15/4, y = (square root equation)
D2 = (-15//4)2 + (square root equation)
Square root equation
If you would be interested in attending a similar session, you can find out more information and apply for our next online Teacher Network meeting on the AMSP website.