A) 12
B) 24
C) \[\frac{1}{4}\]
D) None of these
Correct Answer: B
Solution :
\[ab=2a+3b\] Þ \[(a-3)b=2a\] Þ \[b=\frac{2a}{a-3}\] Now \[z=ab=\frac{2{{a}^{2}}}{a-3}\] Þ \[\frac{dz}{da}=\frac{2[(a-3)2a-{{a}^{2}}]}{{{(a-3)}^{2}}}=\frac{2[{{a}^{2}}-6a]}{{{(a-3)}^{2}}}\] Put \[\frac{dz}{da}=0\], \ \[{{a}^{2}}-6a=0\], \[a=0,\,6\] Now at \[a=6,\]\[\frac{{{d}^{2}}z}{d{{a}^{2}}}=+ve\] When \[a=6,\,\,b=4\]; \ (ab)min. = 6 × 4 = 24.
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