Miguel is designing shipping 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 + 10)(–h + 8). What is the maximum volume for the domain 0 < h < 10? Round to the nearest cubic foot. (1 point) Responses 10 ft3 10 ft 3 107 ft3 107 ft 3 105 ft3 105 ft 3 110 ft3
To find the maximum volume for the given domain, we need to find the maximum value of the function V(h) = h(–h + 10)(–h + 8) when 0 < h < 10.
To do this, we can take the derivative of the function and set it equal to 0 to find the critical points.
V'(h) = (–h + 10)(–h + 8) + h(-1)(-h + 8) + h(-h + 10)(-1)
= (-h + 10)(-h + 8) - h(-h + 8) - h(-h + 10)
= (-h + 10)(-h + 8 - h + 8) - h(-h + 8 - h + 10)
= (-h + 10)(-2h + 16) - h(-2h + 18)
= (-h + 10)(-2h + 16) + h(2h - 18)
= (-h + 10)(-2h + 16 + 2h - 18)
= (-h + 10)(-2)
= 2h - 20
Setting V'(h) = 0:
2h - 20 = 0
2h = 20
h = 10
Since h = 10 is not in the given domain 0 < h < 10, we can ignore it.
Next, we need to check the endpoints of the given domain, 0 and 10.
When h = 0,
V(0) = 0(-0 + 10)(-0 + 8) = 0
When h = 10,
V(10) = 10(-10 + 10)(-10 + 8) = 0
Both endpoints result in a volume of 0.
Therefore, the maximum volume for the given domain 0 < h < 10 is 0 ft³.
To find the maximum volume for the given function V(h) = h(–h + 10)(–h + 8) within the domain 0 < h < 10, we need to find the critical points by taking the derivative of the function and setting it equal to zero.
Let's find the derivative of the function V(h) with respect to h:
V'(h) = 1*(–h + 10)(–h + 8) + h*(-1)(–h + 8) + h(–h + 10)*(-1)
V'(h) = (–h + 10)(–h + 8) - h(–h + 8) - h(–h + 10)
V'(h) = (–h + 10)(–h + 8) + h(h - 8) + h(–h + 10)
V'(h) = (–h + 10)(–h + 8 + h - 8) + h(–h + 10)
V'(h) = (–h + 10)(–h) + h(–h + 10)
V'(h) = h^2 - 10h + 10h - 100
V'(h) = h^2 - 100
Now, set V'(h) equal to zero and solve for h:
h^2 - 100 = 0
(h - 10)(h + 10) = 0
The critical points are h = 10 and h = -10.
Since the domain is restricted to 0 < h < 10, we disregard the negative value h = -10.
So, the only critical point within the given domain is h = 10.
Next, we evaluate the function V(h) at the critical point and the endpoints of the domain:
V(0) = 0(–0 + 10)(–0 + 8) = 0
V(10) = 10(–10 + 10)(–10 + 8) = 10(0)(–2) = 0
V(10) = 0
Comparing the values, we can see that both endpoints and the critical point yield a volume of zero, which means there are no local maximums within the restricted domain of 0 < h < 10.
Therefore, there is no maximum volume within the given domain 0 < h < 10.
The correct response is: 0 ft3
To find the maximum volume for the given function V(h) = h(–h + 10)(–h + 8) within the domain 0 < h < 10, we can use calculus.
First, we need to take the derivative of V(h) with respect to h. Let's call this derivative V'(h):
V'(h) = d/dh [h(–h + 10)(–h + 8)]
To simplify things, we can use the product rule to differentiate the function:
V'(h) = [–h^2 + 18h – 80] + [–h^2 + 10h] + [h(-2h + 10)]
Simplifying further, we get:
V'(h) = –h^2 + 18h – 80 – h^2 + 10h – 2h^2 + 10h
Combining like terms, we have:
V'(h) = –3h^2 + 38h – 80
To find the maximum volume, we need to find the critical points of V(h) by setting V'(h) equal to zero and solving for h:
–3h^2 + 38h – 80 = 0
Now, we can solve this quadratic equation. Factoring may not work easily, so we can use the quadratic formula:
h = (-b ± √(b^2 - 4ac)) / 2a
For this equation, a = -3, b = 38, and c = -80. Substituting the values into the formula, we get:
h = (-38 ± √(38^2 - 4(-3)(-80))) / (2(-3))
Simplifying further, we have:
h = (-38 ± √(1444 - 960)) / -6
h = (-38 ± √484) / -6
h = (-38 ± 22) / -6
Now we have two possible solutions:
1. h = (-38 + 22) / -6 = -16 / -6 = 8/3 ≈ 2.67 (Ignoring this value since it falls outside the given domain)
2. h = (-38 -22) / -6 = -60 / -6 = 10
Since the domain is specified as 0 < h < 10, the only valid solution is h = 10.
Now, we know that the maximum volume occurs at h = 10. We can find this maximum volume by substituting h = 10 into the original volume function V(h):
V(10) = 10(–10 + 10)(–10 + 8)
V(10) = 10(0)(–2)
V(10) = 0
Therefore, the maximum volume of the shipping box within the domain 0 < h < 10 is 0 cubic feet.