A) (10, 10)
B) \[(5,\,\,15)\]
C) (13, 7)
D) None of these
Correct Answer: B
Solution :
\[x+y=20\] and \[z=x{{y}^{3}}\] Þ \[z={{y}^{3}}(20-y)=20{{y}^{3}}-{{y}^{4}}\] Þ \[\frac{dz}{dy}=60{{y}^{2}}-4{{y}^{3}}=0\] Þ \[4{{y}^{2}}(15-y)=0\] So, either \[y=0,\,\] or \[y=15\] Now, \[\frac{{{d}^{2}}z}{d{{y}^{2}}}=120y-12{{y}^{2}}\] ; \ At \[y=0,\ \frac{{{d}^{2}}z}{d{{y}^{2}}}>0\] \\[y=0\]is the point of minima and at \[y=15,\,\,\frac{{{d}^{2}}z}{d{{y}^{2}}}<0\] \ \[y=15\] is the point of maxima. Hence the required parts is (5, 15).
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