Maximum value in enclosed area

Algebra Level 4

Let A A be the set of points in the x y xy -plane such that A = { ( x , y ) 1 x 0 , 0 y x 3 x 2 + x + 1 } . A = \left\{ (x, y) | -1 \leq x \leq 0, 0 \leq y \leq -x^{3}-x^{2}+x+1 \right\} . Then what is the maximum value of 3 x + 4 y ? 3x + 4y ?


The answer is 4.

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1 solution

Tom Engelsman
Oct 29, 2016

The function f(x) = -x^3 - x^2 + x + 1 is strictly increasing over x = [-1, 0]. Hence, the minimum value of 3x + 4y is achieved at (x,y) = (-1,0), or -3. The maximum value is achieved at (x,y) = (0,1), or 4.

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