1 Given that real numbers x and y satisfy the inequality set 2x-y≥0, x+2y≥0 and 3x+y-5≤0, what is the maximum value of 2x+y?

Solution:

This is obviously a linear programming problem,

Most of them have maximum and minimum values at the intersection. Solve three intersections and substitute? There is a maximum at a and a minimum at c;

draw

Three intersections: O(0，0)，A？ ( 1,2),? C(2，- 1)

get

A( 1，2)

Substituting 2x+y has the maximum value.

=2+2=4.

The maximum value of 2x+y is: 4.

2.？ F(x) is a piecewise function,

If so:

The function f(x)={2cos3 derivative x(x≤2000), 2 to the power of x-2008 (x is greater than 2000)},

Then go to 20 13 >: 2000.

get

f(20 13)=2^(20 13-2008)

=2^5

=32

therefore

f[f(20 13)]=f(32)

=2cos32π/3

=2cos2π/3

=2cos(π-π/3)

=-2cosπ/3

=-2×( 1/2)

=- 1.