Given -1 <= x + y <= 4 and 2 <= x - 6y <= 3, find the range of z = 6x - 8y.

Step-by-step solution

1. 1

   Do not try to bound x and y separately. Bounding each variable on its own and recombining always overestimates the range, because the two constraints are linked. The correct method is to express the target as a linear combination of the two given expressions.

2. 2

   Set up the combination. Look for constants $\alpha,\beta$ with

   $6x-8y=\alpha(x+y)+\beta(x-6y).$

   Matching coefficients gives the system

   $\alpha+\beta=6\ (\text{from }x),\qquad \alpha-6\beta=-8\ (\text{from }y).$

3. 3

   Solve for the weights. Subtracting the second equation from the first:

   $7\beta=14\ \Longrightarrow\ \beta=2,\qquad \alpha=6-2=4.$

   Check: $4(x+y)+2(x-6y)=4x+4y+2x-12y=6x-8y$ ✓.

4. 4

   Scale each given interval by its weight. Both weights are positive, so the inequalities keep their direction:

   $-1\le x+y\le 4\ \Longrightarrow\ -4\le 4(x+y)\le 16,$
   $2\le x-6y\le 3\ \Longrightarrow\ 4\le 2(x-6y)\le 6.$

   If a weight had come out negative, that inequality would have to be flipped before adding — the standard trap in this method.

5. 5

   Add the two bounded intervals.

   $-4+4\le z\le 16+6\ \Longrightarrow\ 0\le z\le 22.$

6. 6

   Confirm both endpoints are attainable. The bounds are reached when both constraints sit at their extremes. Taking $x+y=-1$ and $x-6y=2$ gives $x=-\tfrac47$, $y=-\tfrac37$, and $z=6(-\tfrac47)-8(-\tfrac37)=-\tfrac{24}{7}+\tfrac{24}{7}=0$. Taking $x+y=4$ and $x-6y=3$ gives $x=\tfrac{27}{7}$, $y=\tfrac17$, and $z=\tfrac{162}{7}-\tfrac{8}{7}=\tfrac{154}{7}=22$. Both ends are achieved, so the range is exactly $[0,22]$.