- Joined
- Jun 27, 2021
- Messages
- 5,386
- Reaction score
- 422
See attachment. I am skipping part (b).
Part (a)
Volume = length • width • height
V = y • x • x
V = x^2 y
V(x) = x^2 y
I'm not sure about the domain.
Part (c)
The perimeter of the cross section is
P = x + x + x + x + y.
In short, P = 4x + y.
Let P = 108 and solve for y.
108 = 4x + y
108 - 4x = y
Plug 108 - 4x into V(x) for x.
V(x) = x^2 (108 - 4x)
V(x) = 108x^2 - 4x^3
Factor
V(x) = 4x^2(27 - x)
Let V(x) = 0
0 = 4x^2(27 - x)
Set each factor to 0 and solve for x, where x represent the dimension that will maximize the volume.
It is clear that 4x^2 = 0 leads to 0.
So, x = 0 cannot be the answer.
I know 27 - x = 0 leads to 27.
I am going to say x = 27 is the only dimension
for this application that will maximize the volume.
You say?
Part (a)
Volume = length • width • height
V = y • x • x
V = x^2 y
V(x) = x^2 y
I'm not sure about the domain.
Part (c)
The perimeter of the cross section is
P = x + x + x + x + y.
In short, P = 4x + y.
Let P = 108 and solve for y.
108 = 4x + y
108 - 4x = y
Plug 108 - 4x into V(x) for x.
V(x) = x^2 (108 - 4x)
V(x) = 108x^2 - 4x^3
Factor
V(x) = 4x^2(27 - x)
Let V(x) = 0
0 = 4x^2(27 - x)
Set each factor to 0 and solve for x, where x represent the dimension that will maximize the volume.
It is clear that 4x^2 = 0 leads to 0.
So, x = 0 cannot be the answer.
I know 27 - x = 0 leads to 27.
I am going to say x = 27 is the only dimension
for this application that will maximize the volume.
You say?