Using quadratic formula you can have a maximum of two solutions. When you set this problem equal to zero, your a=-1, b=-10, c=2
Answer:
z (min) = 705
x₁ = 10
x₂ = 9
Step-by-step explanation:
Let´s call x₁ quantity of food I ( in ou ) and x₂ quantity of food II ( in ou)
units of vit. C units of vit.E Cholesterol by ou
x₁ 32 9 48
x₂ 16 18 25
Objective function z
z = 48*x₁ + 25*x₂ To minimize
Subject to:
1.-Total units of vit. C at least 464
32*x₁ + 16*x₂ ≥ 464
2.- Total units of vit. E at least 252
9*x₁ + 18*x₂ ≥ 252
3.- Quantity of ou per day
x₁ + x₂ ≤ 35
General constraints x₁ ≥ 0 x₂ ≥ 0
Using the on-line simplex method solver (AtoZmaths) and after three iterations the solution is:
z (min) = 705
x₁ = 10
x₂ = 9
Negative 3.14
Times that by Pi which is 3.14 and then multiply by the base number and you will get the correct answer
The total will be 08.99 because that’s how you divide and he is winning
To solve this problem you must apply the proccedure shown below:
You have the following equation given in the problem above:
<span>-2(bx - 5) = 16
</span> When you solve for bx, you have:
<span>-2(bx - 5) = 16
-2bx-10=16
-2bx=26
bx=26/-2
bx=-13
When you solve for b, you obtain:
</span><span>-2(bx - 5) = 16
-2bx=26
b=-(26/2x)
When yoo solve for x:
</span>2bx=26
x=-(26/2b)<span>
</span>
