Question

The side length of a rectangular box with a square base is increasing at the rate of 2 ft/sec, while the height is decreasing at the rate of 2 ft/sec. At what rate is the volume of the box changing when the side length is 10 ft and the height is 12 ft? (5 points)

680 ft3/sec
40 ft3/sec
-280 ft3/sec
280 ft3/sec

Answer 1

@freckles

Answer 2

@SithsAndGiggles

Answer 3

the base of this box is a L by L in dimensions
since it is a square based box
now the x there symbolizes the height of the box

Answer 4

The volume of a box=length*height*width

Answer 5

our width=length here

Answer 6

\[V=L^2 x\]

Answer 7

the first sentence tells us what about L'?

Answer 8

l increases at 2ft/sec

Answer 9

yeah so we are given L'=2ft/sec

Answer 10

okay

Answer 11

and x'=-2ft/sec

Answer 12

cool

Answer 13

and we are looking for V'

Answer 14

So if V=L^2*x
then V'=?

Answer 15

the first rule you want to apply is product rule

Answer 16

V'=2L(L')(x)+(x')(L^2)

Answer 17

that is beautiful

Answer 18

and we are given L' and x' and we want to find V' when L=12 and x=10

Answer 19

so we just plug in

Answer 20

\[V'=2(12 ft)(2 \frac{ft}{\sec})(10 ft)+(-2 \frac{ft}{\sec})(12 ft)^2\]

Answer 21

V'=280

Answer 22

okay thanks I was just a little overwhelmed when I thought there were a lot of variables

Answer 23

\[V'=480 \frac{ft^3}{\sec}-288 \frac{ft^3}{\sec}=(480-288) \frac{ft^3}{\sec}\]
is that what you did?

Answer 24

I got a different number for V'
but if i made a mistake please correct me

Answer 25

lol

Answer 26

I see I put L=12 and x=10
it was the other way around

Answer 27

\[V'=2(10)(2)(12)-2(10)^2 \\ V'=480-200\]

Answer 28

totally agree with your answer :)

Answer 29

V'=2L(L')(x)+(x')(L^2)
L=10
L'=2
x=12
x'=-2
V'=20(2)(12)+(-2)(100)
V'=480-200=280

Answer 30

ya!

Answer 31

not too terrible right?
:)
just differentiating and pluggin in :)

Answer 32

anyways have a good day