Question: User: 48 a w = -0,01q^3 + 1,5q^2 - 48q - 200
dw/dq = -0,03q^2 + 3q - 48
-0,03q^2 + 3q - 48 = 0
-0,03(q^2 - 100q + 1600) = 0
q^2 - 100q + 1600 = 0
1. **Stating the problem:**
We are given a cubic function $$w = -0.01q^3 + 1.5q^2 - 48q - 200$$ and its derivative $$\frac{dw}{dq} = -0.03q^2 + 3q - 48$$.
We want to find the critical points by solving $$-0.03q^2 + 3q - 48 = 0$$.
2. **Simplify the derivative equation:**
Given:
$$-0.03q^2 + 3q - 48 = 0$$
Divide both sides by $$-0.03$$:
$$\cancel{-0.03}q^2 + \cancel{3}q - \cancel{48} = 0 \div -0.03$$
This gives:
$$q^2 - 100q + 1600 = 0$$
3. **Solve the quadratic equation:**
The quadratic equation is:
$$q^2 - 100q + 1600 = 0$$
Use the quadratic formula:
$$q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $$a=1$$, $$b=-100$$, and $$c=1600$$.
Calculate the discriminant:
$$\Delta = b^2 - 4ac = (-100)^2 - 4 \times 1 \times 1600 = 10000 - 6400 = 3600$$
Calculate the roots:
$$q = \frac{-(-100) \pm \sqrt{3600}}{2 \times 1} = \frac{100 \pm 60}{2}$$
So,
$$q_1 = \frac{100 + 60}{2} = \frac{160}{2} = 80$$
$$q_2 = \frac{100 - 60}{2} = \frac{40}{2} = 20$$
4. **Interpretation:**
The critical points of the function $$w$$ occur at $$q=20$$ and $$q=80$$.
These points can be maxima, minima, or points of inflection depending on the second derivative.
**Final answer:**
$$q = 20 \text{ or } q = 80$$