Solution

The correct answer is 15 and \(\frac{5}{4}\)

EXPLANATION:

The cost function to minimize: Z = 12x + 20y
The nutrient requirements: 6x + 8y ≥ 100, 7x + 12y ≥ 120
The non-negativity constraints: x,y ≥ 0

We use trial and error method, we can substitute the given values for x and y from all four options into the constraint inequalities and validate which of those satisfy both constraints (6x + 8y ≥ 100 and 7x + 12y ≥ 120).

Option 1: 0 and \(12\frac{1}{2}\)

= 6x + 8y ≥ 100

= 6*0 + 8*12/2 ≥ 100

= 0 + 48 ≥ 100

= 48 ≥ 100 it is not greater then or equal to So it is not correct option.

Option 2: 15 and \(\frac{5}{4}\)

= 6x + 8y ≥ 100

= 6*15 + 8* \(\frac{5}{4}\) ≥ 100

= 90 + 20 ≥ 100

= 100 ≥ 100

= 7x + 12y ≥ 120

= 7*15 + 12* \(\frac{5}{4}\) ≥ 120

= 105 + 15  ≥ 120

= 120  ≥ 120

Here both equation satisfy the condition so we not need to check other option.

So correct answer is Option 2: 15 and \(\frac{5}{4}\)