For f(x,y)=xy-2x on the triangular region with vertices (0,0),(4,0),(0,4), evaluating on the edge from (0,4) to (4,0) (line x+y=4, i.e., y=4-x), f becomes a function of x only as:

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SSC Pure Mathematics QUESTION #4095

Question 1

For \(f(x,y)=xy-2x\) on the triangular region with vertices \((0,0),(4,0),(0,4)\), evaluating on the edge from \((0,4)\) to \((4,0)\) (line \(x+y=4\), i.e., \(y=4-x\)), \(f\) becomes a function of \(x\) only as:

\\(f = x(4-x)-2x = -x^2+2x\\)✔️
\(f = 4x-x^2\)
\(f = x(4-x)-2x = -x^2+2x\)
\\(f = 4x-2x = 2x\\)

Correct Answer Explanation

Substitute \(y=4-x\) into \(f=xy-2x\): \(f=x(4-x)-2x=4x-x^2-2x=-x^2+2x\). Maximizing: \(f'=-2x+2=0\Rightarrow x=1\), giving \(f(1,3)=-1+2=1\). This is a candidate for absolute extremum.

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