Q.Refer to Exercise 7 above. Find the maximum value of $Z$.

Exercise 7 sets up $Z=13x-15y$ subject to $x+y\le 7$, $2x-3y+6\ge 0$, $x\ge 0$, $y\ge 0$. Here we want the maximum of $Z$ over that same region.

Corner points of the feasible region:

• $(0,0)$
• $(0,2)$ — where $2x-3y+6=0$ meets $x=0$
• $(3,4)$ — where $x+y=7$ meets $2x-3y+6=0$
• $(7,0)$ — where $x+y=7$ meets $y=0$ …

The referenced problem (Exercise 7)

Exercise 7 asks to minimise $Z=13x-15y$ subject to

This question re-uses the same feasible region but asks for the maximum of $Z$.

Step 1 — Corner points

Boundary lines: $x=0$, $y=0$, $x+y=7$ (intercepts $(7,0),(0,7)$) and $2x-3y+6=0$ (through $(0,2)$ and $(3,4)$).

• $x=0,\;y=0\Rightarrow(0,0)$.
• $x=0$ in $2x-3y+6=0\Rightarrow -3y+6=0\Rightarrow y=2\Rightarrow(0,2)$.
• $x+y=7$ and $2x-3y+6=0$: put $x=7-y$: $2(7-y)-3y+6=0\Rightarrow 20-5y=0\Rightarrow y=4,\,x=3\Rightarrow(3,4)$.
• $x+y=7,\;y=0\Rightarrow(7,0)$. …

Method: Same Region, Opposite Optimum

Use this when a later part re-uses an earlier problem's feasible region but asks for the other optimum (e.g. the maximum where the earlier part found the minimum).

Steps

Step 1: Re-use the same corner points.

The feasible region — and therefore its vertices — is unchanged. Reconstruct (or copy) the same list of corners the earlier part used, keeping only points that satisfy every constraint.

Step 2: Re-read the same $Z$-values. …

Common Mistakes

Mistake 1: Reporting the minimum instead of the maximum.

Why it's wrong: Exercise 7 found the minimum ($-30$ at $(0,2)$) over this region; this part wants the maximum, which is $91$ at $(7,0)$. Copying the earlier answer gives the wrong extreme. Correct approach: from the same corner table, read the largest value for a maximum.

Mistake 2: Including the infeasible point $(0,7)$. …