Question

A rectangular box has a square base with an edge length of x cm and a height of h cm. The volume of the box is given by V = x^2h cm^3. Find the rate at which the volume of the box is changing when the edge length of the base is 12 cm, the edge length of the base is decreasing at a rate of 2 cm/min, the height of the box is 6 cm, and the height is increasing at a rate of 1 cm/min. @freckles

Answer 1

The volume of the box is decreasing at a rate of 144 cm^3/min.
The volume of the box is increasing at a rate of 288 cm^3/min.
The volume of the box is decreasing at a rate of 288 cm^3/min.
The volume of the box is increasing at a rate of 144 cm^3/min.

Answer 2

did you differentiate the equation given w.r.t to time variable t ?

Answer 3

\[V(t)=[x(t)]^2 \cdot h(t)\]
use product rule and chain rule

Answer 4

\[V(t)=2(u)(1)(dh(t)/dx)+h(t) (2)\]

Answer 5

okay I got the same thing

Answer 6

\[V(t)=x^2h \\ V'(t)=2x x' h+x^2h'\]
is this what you mean?

Answer 7

yes

Answer 8

\[V'=? \\ x=12 \\ x'=-2 \\ h=6 \\ h'=1 \]
just enter in the values

Answer 9

and evaluate V'

Answer 10

\[V'=2x x' h+x^2h' \\ V'=2(12)(-2)(6)+(12)^2(1)\]
All I did was just replace the values in the equation with the values I gave me

Answer 11

just use order of operations

Answer 12

V'=-144
@freckles

Answer 13

so the answer is A

Answer 14

24(-12)+144
12(-24+12)
12(-12)
-144
yes seems right
the negative there means the volume is decreasing
and it is decreasing at a rate of 144 cm^3/min

Answer 15

ok great thanks so much!