I'll be perfectly honest, I've never done this type of question before. However, I think I did it right.
Begin by sketching a graph of units sold against price. As it's linear, it's going to be a straight line, crossing the y (units) axis at 140 (when £=0) and the x (price) axis at 70 (when units sold = 0).
The gradient of this line is
ΔxΔy=70−140=−2. So, from the equation of a straight line, using the point (70,0),
y=−2(x−70)=−2x+140.
We want to maximise units x price, that is xy.
xy=x(−2x+140)=−2x2+140x.
This is maximised when its derivative is 0:
dxd(xy)=−4x+140=0, that is when x=35. This can also be found by completing the square, with no calculus:
−2x2+140x=−2(x2−70x)=−2((x−35)2−352) which has a turning point when x=35.
This answer is also what I would intuitively have gone for from the start. Don't put too much reliance on my guessing though - I may be completely wrong.