miquel is designing shiooing boxes that are rectangular prisms the shape of one box with height h in feet has a volume defined by the function v(h)=h(h-5)(h-6) graph the function what is the maximum volume for the domian 0<h(6 round to nearest cubis foot
To graph the function v(h) = h(h-5)(h-6), we can plot some points to generate a rough sketch of the graph.
Let's evaluate v(h) for various values of h within the specified domain, which is 0 < h < 6. We can choose a few values of h and calculate the corresponding values of v(h):
When h = 1: v(1) = 1(1 - 5)(1 - 6) = 1(4)(5) = 20
When h = 2: v(2) = 2(2 - 5)(2 - 6) = 2(-3)(-4) = 24
When h = 3: v(3) = 3(3 - 5)(3 - 6) = 3(-2)(-3) = 18
When h = 4: v(4) = 4(4 - 5)(4 - 6) = 4(-1)(-2) = 8
When h = 5: v(5) = 5(5 - 5)(5 - 6) = 5(0)(-1) = 0
Plotting these points, we would have:
(1, 20), (2, 24), (3, 18), (4, 8), (5, 0)
Now, let's plot these points on a graph:
|
20 |
|
| 24
|
| |
|_________|__________
0 1 2 3 4 5 6
Connecting these points smoothly, we will obtain a graph that increases, reaches a maximum, and then decreases.
To determine the maximum volume for the domain 0 < h < 6, we need to find the highest point on this graph. From the graph, it appears that the maximum volume occurs when h is approximately 2.3.
So, rounding to the nearest cubic foot, the maximum volume for the domain 0 < h < 6 is estimated to be v(2.3) = 24 cubic feet.
To graph the function v(h) = h(h-5)(h-6), we can plot points on a graph to represent different values of h and their corresponding volumes.
First, we can create a table with values of h and the corresponding volumes:
| h | v(h) |
|---|----------|
| 0 | 0 |
| 1 | -4 |
| 2 | -12 |
| 3 | -18 |
| 4 | -20 |
| 5 | -18 |
| 6 | 0 |
Now, let's plot these points on a graph:
- The point (0, 0) represents h = 0 and v(h) = 0.
- The point (1, -4) represents h = 1 and v(h) = -4.
- The point (2, -12) represents h = 2 and v(h) = -12.
- The point (3, -18) represents h = 3 and v(h) = -18.
- The point (4, -20) represents h = 4 and v(h) = -20.
- The point (5, -18) represents h = 5 and v(h) = -18.
- The point (6, 0) represents h = 6 and v(h) = 0.
Now, connect these points with a smooth curve. The resulting graph should look like a parabola or cubic function.
To find the maximum volume within the given domain, we need to find the highest point on the graph of v(h) = h(h-5)(h-6) for the domain 0 < h < 6.
One way to find the maximum volume is by taking the derivative of v(h) with respect to h and setting it equal to zero. So, let's find the derivative:
v'(h) = 3h^2 - 33h + 90
To find the critical points, we set v'(h) = 0 and solve for h:
3h^2 - 33h + 90 = 0
Factoring this quadratic equation, we get:
(h - 6)(3h - 15) = 0
Solving for h, we have two possible critical points:
h = 6 and h = 5
We need to determine which point corresponds to the maximum volume in the given domain. As h = 6 is outside the domain (0 < h < 6), we discard it.
Thus, the only critical point is h = 5.
Now, we calculate the volume at h = 5 to find the maximum volume:
v(5) = 5(5-5)(5-6)
= 5(0)(-1)
= 0
Therefore, the maximum volume within the given domain is 0 cubic feet (rounded to the nearest cubic foot).