Linear Programming and Optimization: Problems with Solutions

Solution:
Let [tex]x_1\ge 0[/tex] and [tex]x_2\ge 0[/tex] be the quantities of [tex]P_1[/tex] and [tex]P_2[/tex] in the mixture. We have to minimize the cost [tex]Z[/tex] subject to the constraints:
[tex]Z=3x_1+2x_2\to\min,\qquad 5x_1+x_2\ge 10,\quad x_1+x_2\ge 6,\quad x_1+2x_2\ge 8[/tex]
Each constraint is the half-plane above its boundary line, so the feasible region is unbounded. Its vertices are the intersections of consecutive boundary lines:
[tex]x_1=0,\ 5x_1+x_2=10\Rightarrow A\left(0 , 10\right)[/tex]
[tex]5x_1+x_2=10,\ x_1+x_2=6\Rightarrow 4x_1=4\Rightarrow B\left(1 , 5\right)[/tex]
[tex]x_1+x_2=6,\ x_1+2x_2=8\Rightarrow x_2=2\Rightarrow C\left(4 , 2\right)[/tex]
[tex]x_1+2x_2=8,\ x_2=0\Rightarrow D\left(8 , 0\right)[/tex]
Each of these points satisfies all constraints (e.g. at [tex]B[/tex]: [tex]x_1+2x_2=11\ge 8[/tex]). Although the region is unbounded, the coefficients of [tex]Z[/tex] are positive, so [tex]Z\ge 0[/tex] is bounded below and its minimum is attained at a vertex:
[tex]Z(A)=20,\quad Z(B)=13,\quad Z(C)=16,\quad Z(D)=24[/tex]
Check without a picture: for every feasible point
[tex]3x_1+2x_2=\frac{1}{4}\left(5x_1+x_2\right)+\frac{7}{4}\left(x_1+x_2\right)\ge\frac{10}{4}+\frac{42}{4}=13[/tex]
The cheapest mixture ([tex]x_1=1[/tex], [tex]x_2=5[/tex]) costs [tex]13[/tex] dollars.

Solution:
The feasible set [tex]D[/tex] is a convex polyhedron; it is non-empty (it contains the origin) and bounded (because [tex]0\le x_i\le 1[/tex]). A linear function is continuous, so on the connected compact set [tex]D[/tex] it takes exactly all values of the closed interval [tex]\left[z_{\min} , z_{\max}\right][/tex].
Since [tex]x_1, x_2, x_3\ge 0[/tex], we have [tex]z\le 0[/tex], and [tex]z=0[/tex] at the feasible point [tex]\left(0 , 0 , 0\right)[/tex]:
[tex]z_{\max}=0[/tex]
From the third constraint [tex]z=-\left(x_1+x_2+x_3\right)\ge -1[/tex]. The point [tex]\left(1 , 0 , 0\right)[/tex] is feasible ([tex]3\le 6[/tex], [tex]1\le 6[/tex], [tex]1\le 1[/tex]) and gives [tex]z=-1[/tex]:
[tex]z_{\min}=-1[/tex]
(The first two constraints are in fact redundant: [tex]3x_1+3x_2+x_3\le 3\left(x_1+x_2+x_3\right)\le 3\lt 6[/tex] and similarly [tex]x_1+x_2+3x_3\le 3\lt 6[/tex].) The set of values is [tex]\left[-1 , 0\right][/tex].

Solution:
At price [tex]x[/tex] the company sells [tex]d(x)=400-10x[/tex] items, which is meaningful for [tex]0\le x\le 40[/tex]. The profit is revenue minus total cost:
[tex]P(x)=x\left(400-10x\right)-10\left(400-10x\right)-2000=\left(x-10\right)\left(400-10x\right)-2000=-10x^2+500x-6000[/tex]
The derivative vanishes at [tex]x=25\in\left[0 , 40\right][/tex], and since [tex]P''(x)=-20\lt 0[/tex] (a downward parabola), this critical point is the global maximum:
[tex]P'(x)=-20x+500=0\Rightarrow x=25,\qquad d(25)=150[/tex]
[tex]P(25)=15\times 150-2000=2250-2000=250[/tex]
The maximum profit is [tex]250[/tex] dollars (at a price of [tex]25[/tex] dollars per item).