2010 Paper 2 Q2 - min / max areas

The diagram shows a triangular piece of land PQR in which angle PQR = 90°, PQ = 8m and QR = 12m. A rectangle QUVW is to be used as the base of a greenhouse, where U, V and W lie on QR, RP and PQ respectively, QU = x m and QW = y m.

(i) Show that 

 

(ii) Express the area A m², of the base of the greenhouse in terms of x.

(iii) Given that x can vary, find the maximum value of A.

(i) Whenever you see a right angled triangle, or any other triangle for that matter, then the use of similar triangles must come to your mind quickly. In this case, we can indeed use similar triangles.

Since PQ = 8m, then PW = 8 – y m

So, using similar triangles,

(ii)  A = xy

A = x (8 – 2x/3)

A = 8x – 2x²/3  m²   (don’t forget your units!!)

(iii) Although we could use differentiation to find the maximum value of A, we note that this is a quadratic function, so we can just use completing the square to get it.

Since this is a negative x² function, it means that there is a max area.

Using completing the square method:

A = ⅔(12x – x²)

   = -⅔[(x – 6)² – 36]

   = -⅔(x – 6)² + 24

Hence, the max area is 24 m² and occurs when x = 6m