A) 20
B) 16
C) 24
D) 8
E) 4
Correct Answer: D
Solution :
\[x+2y=8\], \[y=\frac{8-x}{2}\] Now \[f(x)=xy=x.\frac{(8-x)}{2}=4x-\frac{{{x}^{2}}}{2}\] \[\therefore \] \[{f}'(x)=4-x\] For extremum,\[{f}'(x)=0\] \[\therefore \] \[x=4\] and y = 2. Also \[{f}''(x)=-1<0\] So, maximum value of \[xy=4\times 2=8\].You need to login to perform this action.
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