Answer:
[2 5 0]
[-7 -11/2 12]
Step-by-step explanation:
We want to perform elementary row operation represented by R2 - ½R1 on matrix A in the figure attached.
Now, matrix A is given as;
[2 5 0]
[-6 -3 12]
Now, R1 is row 1 = [2 5 0]
R2 is row 2 = [-6 -3 12]
Thus, R2 - ½R1 = [-6 -3 12] - ½[2 5 0]
This gives;
[-6 -3 12] - [1 5/2 0] = [-7 -11/2 12]
Now,[-7 -11/2 12] will be the new row 2 since it's just an elementary row operation we did on the Matrix A.
Thus, the new matrix is now;
[2 5 0]
[-7 -11/2 12]
Answer:
Step-by-step explanation:
Let
and
be the production level of milk and white chocolate-covered strawberries respectively. According to the given data, we know the total profit will be

The restrictions can be written as



All the restrictions can be plotted in the same graph to find the feasible region where all of them are met. The graph is shown in the image below
The optimal solution will be the level of production such that
* All restrictions are met
* The total profit is maximum
The optimal level of production can be found in (at least) one of the vertices of the feasible region. We'll try each one as follows
P(0,0)=0
P(400,200)=$2.25 (400)+$2.50 (200) = $1400
P(600,200)=$2.25 (600)+$2.50 (200) = $1850
P(800,0)=$2.25 (800)+$2.50 (0) = $1800
We must produce 600 milk chocolate-covered strawberries and 200 white chocolate-covered strawberries to have a maximum profit of $1850/month
Answer:
10
Step-by-step explanation:
(1 1/4) * 8 = 10
Answer: 2.852 × 10-1
Step-by-step explanation:
7.13 x 10^-3 = 0.00713
0.00713 x 40 = 0.2852
0.2852 = 2.852 × 10-1