The unshaded region on the Cartesian plane satisfies the inequalities; y >= -1 1/2x + 5 1/2, y < 1/2x + 3. Find the minimum value of (x+2y) for the integral coordinates R(x,y) lying in the unshaded region.

• Mstari wa kwanza (L1): $y = -\frac{3}{2}x + \frac{11}{2}$. Huu ni mstari imara (solid line) kwa sababu ya usawa $\ge$. Eneo lisilo na kivuli liko juu au kwenye mstari huu.
• Mstari wa pili (L2): $y = \frac{1}{2}x + 3$. Huu ni mstari wenye nukta (dashed line) kwa sababu ya usawa $<$. Eneo lisilo na kivuli liko chini ya mstari huu.
• Mstari wa tatu (L3): $y = \frac{3}{2}x + \frac{3}{2}$. Huu ni mstari wenye nukta (dashed line) kwa sababu ya usawa $<$. Eneo lisilo na kivuli liko chini ya mstari huu.

Eneo lisilo na kivuli ni pembetatu ndogo iliyoko katikati ya grafu, iliyofungwa na mistari hii mitatu.

Tunahitaji kupata pointi ambapo $x$ na $y$ zote ni namba kamili na zinakidhi usawa wote watatu.
Hebu tuchunguze pointi za gridi zinazowezekana karibu na eneo hili.

Kwa $x=2$:

1. $y \ge -\frac{3}{2}(2) + \frac{11}{2} \implies y \ge -3 + 5.5 \implies y \ge 2.5$
2. $y < \frac{1}{2}(2) + 3 \implies y < 1 + 3 \implies y < 4$
3. $y < \frac{3}{2}(2) + \frac{3}{2} \implies y < 3 + 1.5 \implies y < 4.5$ Kwa $x=2$, tunahitaji $2.5 \le y < 4$. Namba kamili ya $y$ inayokidhi hili ni $y=3$. Hivyo, pointi $(2,3)$ inakidhi usawa wote.

Kwa $x=3$:

1. $y \ge -\frac{3}{2}(3) + \frac{11}{2} \implies y \ge -4.5 + 5.5 \implies y \ge 1$
2. $y < \frac{1}{2}(3) + 3 \implies y < 1.5 + 3 \implies y < 4.5$
3. $y < \frac{3}{2}(3) + \frac{3}{2} \implies y < 4.5 + 1.5 \implies y < 6$ Kwa $x=3$, tunahitaji $1 \le y < 4.5$. Namba kamili za $y$ zinazokidhi hili ni $y=1, 2, 3, 4$. Hivyo, pointi $(3,1)$, $(3,2)$, $(3,3)$, $(3,4)$ zinakidhi usawa wote.

Kwa $x=4$:

1. $y \ge -\frac{3}{2}(4) + \frac{11}{2} \implies y \ge -6 + 5.5 \implies y \ge -0.5$
2. $y < \frac{1}{2}(4) + 3 \implies y < 2 + 3 \implies y < 5$
3. $y < \frac{3}{2}(4) + \frac{3}{2} \implies y < 6 + 1.5 \implies y < 7.5$ Kwa $x=4$, tunahitaji $-0.5 \le y < 5$. Namba kamili za $y$ zinazokidhi hili ni $y=0, 1, 2, 3, 4$. Hivyo, pointi $(4,0)$, $(4,1)$, $(4,2)$, $(4,3)$, $(4,4)$ zinakidhi usawa wote.

Pointi zote za kuratibu kamili katika eneo lisilo na kivuli ni:
$(2,3)$, $(3,1)$, $(3,2)$, $(3,3)$, $(3,4)$, $(4,0)$, $(4,1)$, $(4,2)$, $(4,3)$, $(4,4)$.

• Kwa $(2,3)$: $x+2y = 2 + 2(3) = 2 + 6 = 8$
• Kwa $(3,1)$: $x+2y = 3 + 2(1) = 3 + 2 = 5$
• Kwa $(3,2)$: $x+2y = 3 + 2(2) = 3 + 4 = 7$
• Kwa $(3,3)$: $x+2y = 3 + 2(3) = 3 + 6 = 9$
• Kwa $(3,4)$: $x+2y = 3 + 2(4) = 3 + 8 = 11$
• Kwa $(4,0)$: $x+2y = 4 + 2(0) = 4 + 0 = 4$
• Kwa $(4,1)$: $x+2y = 4 + 2(1) = 4 + 2 = 6$
• Kwa $(4,2)$: $x+2y = 4 + 2(2) = 4 + 4 = 8$
• Kwa $(4,3)$: $x+2y = 4 + 2(3) = 4 + 6 = 10$
• Kwa $(4,4)$: $x+2y = 4 + 2(4) = 4 + 8 = 12$

Thamani ndogo zaidi kati ya hizi ni 4.

Thamani ndogo zaidi ya $(x+2y)$ ni $\boxed{4}$.