Example

Solving a Marathon Calorie and Budget Problem Using a System of Inequalities

Problem: Tenison needs to eat at least an extra 1{,}000 calories a day to prepare for a marathon. He has only $25 to spend on the extra food and will spend it on donuts (costing $0.75 and having 360360 calories each) and energy drinks (costing $2 and having 110110 calories each).

ⓐ Set up the system. Let dd = the number of donuts and ee = the number of energy drinks. Translating the two constraints:

  • "At least 1{,}000 calories" → 360d+110e≥1,000360d + 110e \geq 1{,}000
  • "No more than $25" → 0.75d+2e≤250.75d + 2e \leq 25

Because quantities cannot be negative, we also have d≥0d \geq 0 and e≥0e \geq 0. The system is: {360d+110e≥1,0000.75d+2e≤25d≥0e≥0\left\{\begin{array}{l} 360d + 110e \geq 1{,}000 \\ 0.75d + 2e \leq 25 \\ d \geq 0 \\ e \geq 0 \end{array}\right.

ⓑ Graph the system. Graph 360d+110e=1,000360d + 110e = 1{,}000 as a solid boundary line. Testing (0, 0): 0≥1,0000 \geq 1{,}000 is false, so shade the side away from the origin. Graph 0.75d+2e=250.75d + 2e = 25 as a solid boundary line. Testing (0, 0): 0≤250 \leq 25 is true, so shade the side containing the origin. The solution is the doubly-shaded region in the first quadrant.

ⓒ To determine if 88 donuts and 44 energy drinks satisfy the needs, we test the point (8, 4) in the inequalities:

  • Calories: 360(8)+110(4)=2,880+440=3,320≥1,000360(8) + 110(4) = 2{,}880 + 440 = 3{,}320 \geq 1{,}000 (True)
  • Budget: 0.75(8)+2(4)=6+8=14≤250.75(8) + 2(4) = 6 + 8 = 14 \leq 25 (True) Since both constraints are met, he can buy 88 donuts and 44 energy drinks.

ⓓ To determine if 11 donut and 33 energy drinks satisfy the needs, we test the point (1, 3):

  • Calories: 360(1)+110(3)=360+330=690≥1,000360(1) + 110(3) = 360 + 330 = 690 \geq 1{,}000 (False) Since the calorie constraint is not met, he cannot buy 11 donut and 33 energy drinks.

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Updated 2026-06-03

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