Question: Consider the following one-dimensional problem: This is a concave problem, since the leading term in the quadratic objective is negative, so that the second-order derivative

Consider the following one-dimensional problem:

min (x-2)+3 s.t. 1 <

This is a concave problem, since the leading term in the quadratic objective is negative, so that the second-order derivative is negative everywhere. In Fig. 15.5, we show the objective function and the feasible set. The stationarity point x = 2 is of no use to us, since it is a maximizer. We see that local minimizers are located at the boundary of the feasible set. A local minimizer lies at the left boundary, x = 1, and the global minimizer is located at the right boundary, x = 4.

Data From Fig 15.5

image text in transcribed

min (x-2)+3 s.t. 1 <

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