A diet is to contain at least 2270 units of carbohydrates, 2378 units of protein

Question

A diet is to contain at least 2270 units of carbohydrates, 2378 units of proteins, and 1872 calories. Two foods are available: F1 which costs $ 0. 03 per unit and F2, which costs $ 0. 01 per unit. A unit of food F1 contains 5 units of carbohydrates, 7 units of proteins and 8 calories. A unit of food F2 contains 5 units of carbohydrates, 3 units of proteins and 2 calories. Find the minimum cost for a diet that consists of a mixture of these two foods and also meets the minimal nutrition requirements. Corner points of the feasible region: If there is more than one corner point, type the points separated by a comma (i. e. (1, 2), (3, 4)). Minimum cost is: when F1 = and F2 =

Explanation / Answer

Let X =F1, Y = F2
The constrains are:
(1) 5X+5Y>=2270
(2) 7X+2Y>=2378
(3) 8X+2Y>=1872
X>=0, Y>=0


If we solve Equation 1 and 2, we will get: x = 254, y = 200

If we solve Equation 2 and 3, we will get: x = 160.67, y = 293.33 (We can eliminate this as this will not fulfil the second constraints)

If we solve Equation 1 and 3, we will get: x = 86, y = 592


Thus, corner of feasible region = (254,200) (86,592)
The objective is to minimise cost.
Thus,
Based on F1=254 units, F2=200 units, cost will be 254 x 0.03 + 200 x 0.01 = $9.62
Based on F1=86 units, F2=592 units, cost will be 86 x 0.03 + 592 x 0.01 = $8.50

Thus, Minimum cost is $8.50. F1= 86 units; F2 = 592 units

Hope this helps!

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