To analyze the equation \( A(w) = w^2 + 5w \), we can rewrite it in a standard quadratic form:
\[ A(w) = w^2 + 5w \]
This is a quadratic equation where the coefficient of \( w^2 \) is positive (1), indicating that the parabola opens upwards.
To find the vertex of the parabola, which represents the maximum or minimum of the function, we can use the formula for the vertex \( w = -\frac{b}{2a} \), where \( a = 1 \) and \( b = 5 \):
\[ w = -\frac{5}{2 \cdot 1} = -\frac{5}{2} \]
Since this vertex is at \( w = -2.5 \) (which is outside the range we are interested in since \( w \) must be greater than 0 for the context of the problem), we conclude that:
- The function has no maximum value since it continues to increase as \( w \) increases.
- The area \( A(w) \) does not have a maximum at \( w = 5 \), and it cannot have a minimum at that value either.
Now, looking at the possible options:
- A(w) has a maximum value at w=5. (Incorrect, it has no maximum)
- A(w) has a minimum value at w=5. (Incorrect)
- A(w) increases as w increases for 0<w<5. (Correct, because the function is an upward-opening parabola)
- A(w) increases as w increases when w>0. (Correct, because the function is always increasing for \( w > -2.5 \))
Therefore, the most accurate choices are:
- A(w) increases as w increases for 0<w<5.
- A(w) increases as w increases when w > 0.
However, if you must choose only one, the best answer that specifically highlights the behavior over the stated interval is:
A(w) increases as w increases for 0<w<5.