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**engacenus** · 04-05-2011, 07:40 PM

$\\\pi_1=TR_1-TC_1=\frac{(2a - \lambda_1)^2}{36b}=\frac{(150 - \lambda_1)^2}{72} \\\\\pi_2=TR_2-TC_2=\frac{(a - 2\lambda_1)^2}{9b}=\frac{(75 - 2\lambda_1)^2}{18}$

a=75; b=2

That LaTeX site is brilliant! If only that existed when I needed it for short equations.

I agree your answers, didn't at first because when I did Q1.Q2 I put (-)lambda^2 instead of (+)lambda^2.

04-05-2011, 07:40 PM	#12
engacenus Join Date Oct 2005 Posts 494 Senior Member	$\\\pi_1=TR_1-TC_1=\frac{(2a - \lambda_1)^2}{36b}=\frac{(150 - \lambda_1)^2}{72} \\\\\pi_2=TR_2-TC_2=\frac{(a - 2\lambda_1)^2}{9b}=\frac{(75 - 2\lambda_1)^2}{18}$ a=75; b=2 That LaTeX site is brilliant! If only that existed when I needed it for short equations. I agree your answers, didn't at first because when I did Q1.Q2 I put (-)lambda^2 instead of (+)lambda^2.
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