All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

4.
(i)
By Pythagoras’ Theorem, (2x)^{2}  + (y-2)^{2} = (2\sqrt{65})^{2}

\therefore, 5x^{2} + y^{2}-2xy = 260

(ii)

Perimeter = 6x+2y = 60

\Rightarrow 5x^{2} + (30-3x)^{2}-2x(30-3x) = 260

x^{2} -12x +32 = 0

x = 4 ~\mathrm{or}~ x= 8

When x = 4, y = 18

When $latex x = 8, y = 6 < x$, which is not possible given the question. Thus, rejected. $latex \therefore, x=4, y=18$

KS Comments

Students need to be alert to reject the appropriate x-value. Alternatively, they can also check the second order derivative to determine which is the suitable x value.

One Comment

Leave a Reply