Calculating Constant Volume Rate of Change

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SUMMARY

The discussion focuses on calculating the rate of change of width for a rectangular object with a fixed length of 1 meter, maintaining a constant volume of 16,000 cm³ while the height increases at 12 cm/min. The user correctly applies the formula for volume, V = length × width × height, and derives the relationship between the rates of change of the dimensions. The calculated rate of change for width (dz/dt) is -7.5 cm/min, indicating that as height increases, width must decrease to keep the volume constant.

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Homework Statement


A rectangular object has a fixed length of 1m. The height is increasing by 12 cm/min. Find the rate that the width must change so that the volume remains constant at 16 000cm^3 when the height is 10 cm





The Attempt at a Solution



So here's what I tried:

The Volume= Lenth x Width x Height
V = x y z Since x is fixed at 100 , V= 100 y z
dV/dt = 100y dz/dt +100 z dy/dt = 0
z dy/dt = -ydz/dt
(10) (12) = - y dz/dt
When x= 100 and z= 10 and V=16000 , y = 16
dz/dt = - (10)(12)/y = - 120/16 = -7.5 cm/min


However, I don't even know if I am taking the right approach... Please give me assistance
 
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Stanc said:

Homework Statement


A rectangular object has a fixed length of 1m. The height is increasing by 12 cm/min. Find the rate that the width must change so that the volume remains constant at 16 000cm^3 when the height is 10 cm





The Attempt at a Solution



So here's what I tried:

The Volume= Lenth x Width x Height
V = x y z Since x is fixed at 100 , V= 100 y z
dV/dt = 100y dz/dt +100 z dy/dt = 0
z dy/dt = -ydz/dt
(10) (12) = - y dz/dt
When x= 100 and z= 10 and V=16000 , y = 16
dz/dt = - (10)(12)/y = - 120/16 = -7.5 cm/min


However, I don't even know if I am taking the right approach... Please give me assistance

Looks OK to me, but I didn't check that closely. It's reasonable to get a negative rate, since one dimension is increasing, and one is constant. It has to be true that the third dimension is decreasing, this you get a negative rate.

It would have been helpful to use variables that matched what they represent - h for height, and w for width. I have to do a bit of translation with y and z.
 
Mark44 said:
Looks OK to me, but I didn't check that closely. It's reasonable to get a negative rate, since one dimension is increasing, and one is constant. It has to be true that the third dimension is decreasing, this you get a negative rate.

It would have been helpful to use variables that matched what they represent - h for height, and w for width. I have to do a bit of translation with y and z.

Ok sorry about the representation thing but is my approach correct? I just don't understand why I didnt have to use the chain rule...
 

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