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vincisonfire · #2 December 1st, 2008, 04:36 PM

(i) Let x and y be the cathete lengths.
Condition : $x+y=12 \implies y = 12 -x$
$H^2 =x^2+y^2 = x^2 + (12-x)^2$
$2H\frac{dH}{dx} =2x-2(12-x) = 4x-24 =0 \implies x =6$
If you maximise $H^2$ you also maximise $H$
(ii) Same principle
Let x be the little side and y be the long side
Condition : $2x+2y=l \implies y = \frac{l}{2} -x$
$S =xy = x\frac{l-2x}{2}$
You differentiate and set equal to 0.
Use the second derivative test to see the nature of the extrema .