The anwser to your problem is 1,800
Answer:
The correct option is;
B
Step-by-step explanation:
The given system of inequalities are;
5·x - 4·y > 4...(1)
x + y < 2...(2)
Representing both inequalities as a function of "y", gives;
For, 5·x - 4·y > 4...(1), we have;
-4·y > 4 - 5·x
y < 4/(-4) - 5·x/(-4)
∴ y < 5·x/4 - 1
For x + y < 2...(2), we have;
y < 2 - x
Therefore, y is less than the values given by the equation of the straight line equalities, and the feasible region is given by the common region under both dashed lines representing both inequalities as shown in the attached diagram created using Microsoft Excel
The correct option is therefore, B.
Because we know the area in terms of paint flow, and paint flow in terms of time, we can substitute p(t) for p in the A(p) equation.
A(p(t)) = A(t) = <span>π * (5t)^2 (assuming it's squared for the A(p).
B: 314 units^2
If A(p) = </span>πp2 (instead of p^2), then A(t) = 10πt
B: 31.4 units^2