the height of a rectangular box is 10 in. Its length increases at the rate of 2 in/sec.;its width decreases at the rate of 4 in/sec. When the length is 8 in and the width is 6 in, the volume of the box is changing in cubic inches per second at the rate of ?
A 200
B 80
C -80
D -200
E -20
v = 10xy
dv/dt = 10(y dx/dt + x dy/dt)
So, using your numbers,
dv/dt = 10(8(-4) + 6(2)) = -200
looks good to me.
ty!
i got -200 but wasn't sure
To find the rate at which the volume of the box is changing, we can use the formula for the volume of a rectangular box:
Volume = Length x Width x Height
Given:
Height = 10 in
Length increasing rate = 2 in/sec
Width decreasing rate = 4 in/sec
We need to find the rate of change of the volume when the length is 8 in and the width is 6 in.
Let's calculate the rate of change of the volume:
First, substitute the given values into the volume formula:
Volume = (Length) x (Width) x (Height)
Volume = (8 in) x (6 in) x (10 in)
Volume = 480 in³
Next, differentiate the volume with respect to time (t):
dV/dt = (dL/dt) x (W x H) + (L x dW/dt) x (H) + (L x W) x (dH/dt)
Since we are given that the length is increasing at the rate of 2 in/sec and the width is decreasing at the rate of 4 in/sec, we can substitute these values into the derivative equation:
dV/dt = (2 in/sec) x (6 in x 10 in) + (8 in x -4 in/sec) x (10 in) + (8 in x 6 in) x (0)
Simplifying:
dV/dt = (2) x (60) + (-32) x (10) + (48) x (0)
dV/dt = 120 + (-320) + 0
dV/dt = -200 in³/sec
Therefore, the rate of change of the volume of the box is -200 cubic inches per second.
The correct answer is D: -200.
To find the rate at which the volume of the box is changing, we can use the product rule of differentiation.
The volume of a rectangular box is given by V = length × width × height.
Given information:
- Height (h) = 10 in
- Rate of change of length (dl/dt) = 2 in/sec
- Rate of change of width (dw/dt) = -4 in/sec
- Length (l) = 8 in
- Width (w) = 6 in
We need to find the rate of change of volume (dV/dt) when l = 8 in and w = 6 in.
Let's differentiate the volume equation with respect to time (t):
dV/dt = (d/dt)(length × width × height)
To differentiate a product of functions, we use the product rule: d(uv)/dt = u * dv/dt + v * du/dt
In this case:
u = length (l)
v = width (w)
du/dt = dl/dt (rate of change of length)
dv/dt = dw/dt (rate of change of width)
Applying the product rule:
dV/dt = l * dw/dt * h + w * dl/dt * h
Substituting the given values:
dV/dt = (8) * (-4) * (10) + (6) * (2) * (10)
Simplifying:
dV/dt = -320 + 120
dV/dt = -200 cubic inches per second
Therefore, the rate of change of the volume of the box is -200 cubic inches per second.
The correct answer is D) -200.
Well, well, we have a rectangular box that is having a lot of fun with its dimensions! Let's see what's happening here.
First, we need to calculate the rate of change of the volume with respect to time.
The volume of a rectangular box is given by the equation V = lwh, where l is the length, w is the width, and h is the height.
Now, we know that the height of the box is constant at 10 inches, so we can ignore it for now.
The rate of change of the volume with respect to time can be calculated using the chain rule of calculus.
δV/δt = (δV/δl) * (dl/dt) + (δV/δw) * (dw/dt)
Here, (δV/δl) and (δV/δw) represent the partial derivatives of the volume with respect to l and w, respectively.
Given that l is increasing at a rate of 2 in/sec and w is decreasing at a rate of 4 in/sec, we can substitute these values in the above equation.
δV/δt = (10 * 2) + (10 * -4)
Simplifying the equation, we get:
δV/δt = 20 - 40
Voila! The rate of change of the volume of the box is -20 cubic inches per second. So, my dear friend, the answer is E) -20.
But don't be too hard on the box, it's just trying to shrink and grow at the same time!