Since 48% of students are in favor of the change, we need to find 48% of 720 students:

0.48 * 720 = 345.6

However, since we can't have a fraction of a student, we round to the nearest whole number. Thus, approximately 346 students are in favor of the change.

Since a cube has 6 congruent square faces, the area of each face is equal to the total surface area divided by 6. So, the area of each face is 150 / 6 = 25 square units.

Let s be the length of a side of the cube. Each face is a square, so its area is also equal to s². Therefore, s² = 25 square units, and we find that s = 5 units by taking the square root of 25.

The volume of a cube is given by V = s³, so in this case, the volume is 5³ = 125 cubic units.

Let x be the number of Widget A units and y be the number of Widget B units. We have the following inequalities:

5x + 7y ≥ 8000 (profit constraint)
x + y ≤ 1200 (production constraint)

To minimize the number of Widget A units, we want to maximize the number of Widget B units. We can rewrite the production constraint as:

x ≥ 1200 - y

Substituting this into the profit constraint, we get:

5(1200 - y) + 7y ≥ 8000

Simplifying and solving for y, we have:

6000 - 5y + 7y ≥ 8000
2y ≥ 2000
y ≥ 1000

Since we want to minimize x, we can take the maximum possible value for y:

y = 1000

Now, we find the value of x:

x ≥ 1200 - y = 1200 - 1000 = 200

So, the minimum number of Widget A units the company must sell is 200.

To find the value of y:

Divide both sides of the equation by 7:
y = 63 / 7
y = 9

So, y equals 9.

The profit made for each item is $3, so the total daily profit made from x items is 3x.

Since the factory operates for 15 days, the total profit after 15 days is:

Total Profit = 15 * daily profit = 15 * 3x = 45x

Therefore, the factory will make a profit of 45x dollars after 15 days.