1. **Problem Statement:** Mary wants to maximize her profit by baking two types of cakes: Special and Standard. She makes 70 profit per Special cake and 50 profit per Standard cake. She has 500 Kg of flour and each Special cake requires 2.5 Kg, each Standard cake requires 1.5 Kg. She can bake at most 300 cakes per weekend. 2. **Formulating the Linear Programming Problem:** Let $x$ = number of Special cakes, $y$ = number of Standard cakes. Objective function (maximize profit): $$\text{Maximize } Z = 70x + 50y$$ Subject to constraints: - Flour constraint: $$2.5x + 1.5y \leq 500$$ - Cake quantity constraint: $$x + y \leq 300$$ - Non-negativity: $$x \geq 0, y \geq 0$$ 3. **Finding the Optimal Mix:** We solve the system by checking corner points of the feasible region. - From $$x + y = 300$$, express $$y = 300 - x$$. - Substitute into flour constraint: $$2.5x + 1.5(300 - x) \leq 500$$ $$2.5x + 450 - 1.5x \leq 500$$ $$1.0x \leq 50$$ $$x \leq 50$$ So feasible $$x$$ values are between 0 and 50. Check corner points: - Point A: $$x=0, y=0$$ $$Z=70(0)+50(0)=0$$ - Point B: $$x=0, y=300$$ (check flour: $$2.5(0)+1.5(300)=450 \leq 500$$ valid) $$Z=70(0)+50(300)=15000$$ - Point C: $$x=50, y=250$$ (since $$x+y=300$$) Check flour: $$2.5(50)+1.5(250)=125+375=500$$ valid $$Z=70(50)+50(250)=3500+12500=16000$$ - Point D: $$x=200, y=0$$ (check flour: $$2.5(200)+1.5(0)=500$$ valid but $$x+y=200 \leq 300$$) $$Z=70(200)+50(0)=14000$$ 4. **Conclusion:** The maximum profit is at Point C with $$x=50$$ Special cakes and $$y=250$$ Standard cakes, yielding a profit of $$16000$$. Hence, Mary should bake 50 Special cakes and 250 Standard cakes to maximize her profit.