My solution is based on 3-dimensional graphics. The three inequalities establish the borders and vertices of a solid. Four of the vertices lie on the axes as intercepts in a 3D Cartesian reference frame. One of the vertices is the origin O(0,0,0). To find the other 3 vertices (intercepts) we can zeroise two variables in three equations derived by simply removing the inequality signs and replacing them with equals. Then it is easy to find the intercepts, which I’ll label with letters A-H. The intercepts are the constants divided by the coefficients of the non-zeroed variable.
|
x₁
|
x₂
|
x₃
|
Equation
|
|
A(5/2,0,0)
|
D(0,10/3,0)
|
G(0,0,20)
|
8x₁+6x₂+x₃=20
|
|
B(5,0,0)
|
E(0,10,0)
|
H(0,0,40/3)
|
4x₁+2x₂+1.5x₃=20
|
|
C(4,0,0)
|
F(0,16/3,0)
|
I(0,0,16)
|
2x₁+1.5x₂+0.5x₃=8
|
By joining the intercepts on each plane we get 9 boundary lines:
AD, AG, DG, BE, BH, EH, CF, CI, FI. From these we select the boundary lines closest to the origin. In the x₁-x₂ plane, the boundary line AD is closest to the origin because, for x₁, x₂≥0, none of the boundary lines AD, BE, CF intersect.
However, in the x₁-x₃ plane the boundary lines do intersect, as do those in the x₂-x₄ plane. So we know that the endpoints of AD will be involved, and we can discard BE and CF.
Now we focus on the intersection of AG with BH and CI, and DG with EH and FI. When they intersect, these lines will create vertices of the solid figure. We have to find out which vertices are closer to the origin O. Then we will have optimised the constraints given by the inequalities. In other words, all the inequalities will be satisfied.
We have 4 systems of equations to solve to find possible vertices. We know we only have to deal with two planes, one for each pair of equations.
Let’s look at the intersection of AG and BH first. x₂=0, so the equations for AG and BH reduce to:
8x₁+x₃=20, so x₃=20-8x₁, and 4x₁+1.5x₃=20.
Substituting for x₃: 4x₁+1.5(20-8x₁)=20,
4x₁+30-12x₁=20, 10=8x₁, x₁=5/4, making x₃=20-10=10.
So there’s a vertex J(5/4,0,10).
Next, AG with CI.
We have x₃=20-8x₁ already. The other equation is:
2x₁+0.5x₃=8. Substituting for x₃:
2x₁+10-4x₁=8, 2=2x₁, x₁=1, making x₃=12.
Call that vertex K(1,0,12).
Which vertex is nearer O? To find out we measure the distance from O radially:
OJ²=(5/4)²+10²=101.5625;
OK²=1²+12²=145. Therefore OJ is nearer and J(5/4,0,10) is the required vertex.
Next, consider DG with EH. This time x₁=0 so the equations are:
6x₂+x₃=20, so x₃=20-6x₂, and 2x₂+1.5x₃=20.
2x₂+1.5(20-6x₂)=20, 2x₂+30-9x₂=20,
10=7x₂, x₂=10/7, making x₃=20-60/7=80/7.
Vertex L(0,10/7,80/7).
Now, DG with FI.
x₃=20-6x₂ and 1.5x₂+0.5x₃=1.5x₂+10-3x₂=8,
2=1.5x₂, x₂=4/3, making x₃=12.
Vertex M(0,4/3,12).
OL²=(100/49)(1+64)=6500/49=132.65;
OM²=16/9+144=145.78.
So L(0,10/7,80/7) is the closer vertex.
Now that we have all 6 vertices we can now apply the formula:
60x₁+30x₂+20x₃ to each certified vertex:
O=0 (that’s the minimum),
A(5/2,0,0)=150, D(0,10/3,0)=100, H(0,0,40/3)=800/3=266.67,
J(5/4,0,10)=275, L(0,10/7,80/7)=1900/7=271.43.
The maximum of these is 275, corresponding to:
x₁=5/4 (1.25), x₂=0, x₃=10.
Finally we need to check that this set satisfies the given constraints.
8x₁+x₃=20, 4x₁+1.5x₃=20, 2x₁+0.5x₃=7.5 (<8) OK!