The problem provides a function $h(v) = -v + 3000 - \frac{8100}{v}$ that models the daily production of a machine, where $v$ is the speed in km/h and $v \in [50, 120]$. The first question asks to determine the speed $v$ that maximizes the daily production $h(v)$, and then calculate the maximum daily production.

1. Problem Description

The problem provides a function $h(v) = -v + 3000 - \frac{8100}{v}$ that models the daily production of a machine, where $v$ is the speed in km/h and $v \in [50, 120]$. The first question asks to determine the speed $v$ that maximizes the daily production $h(v)$, and then calculate the maximum daily production.

2. Solution Steps

To find the maximum of $h(v)$, we first find the derivative $h'(v)$ and then set it equal to zero to find the critical points.

$h(v) = -v + 3000 - \frac{8100}{v}$

$h'(v) = -1 + \frac{8100}{v^2}$

To find critical points, we set $h'(v) = 0$:

$-1 + \frac{8100}{v^2} = 0$

$\frac{8100}{v^2} = 1$

$v^2 = 8100$

$v = \pm \sqrt{8100} = \pm 90$

Since $v \in [50, 120]$, we only consider the positive solution $v = 90$.

Now, we need to verify that $v = 90$ indeed gives a maximum. We can find the second derivative $h''(v)$:

$h'(v) = -1 + 8100v^{-2}$

$h''(v) = -16200v^{-3} = -\frac{16200}{v^3}$

Since $v = 90$, $h''(90) = -\frac{16200}{90^3} = -\frac{16200}{729000} < 0$.

Because the second derivative is negative, we have a maximum at $v = 90$.

Now we calculate the maximum daily production by substituting $v = 90$ into $h(v)$:

$h(90) = -90 + 3000 - \frac{8100}{90} = -90 + 3000 - 90 = 2820$

3. Final Answer

The speed that maximizes the daily production is $v = 90$ km/h. The maximum daily production is $2820$ units.