We want to find the rate of change of the surface for a cube of increasing volume.
When the side length is equal to 20 inches, the surface area is increasing at a rate of 439.29 in^2 per minute.
We know that for a cube of side length S, the volume is:
V = S^3
And the surface area is:
A = 6*S^2
Here we know that the volume increases at a rate of 12in^3/min.
Because the volume depends on the side length cubed, we can say that the side length increases at a rate of:
∛(12in^3)/min = 2.29 in/min
Then, as a function of time t in minutes, we can write the side length as:
S(t) = (2.29 in/min)*t
Then the surface area equation is:
A(t) = 6*((2.29 in/min)*t)*((2.29 in/min)*t)
A(t) = (31.46 in^2/min^2)*t^2
The rate of change is given by direct differentiation, we get:
A'(t) = 2*(31.46 in^2/min^2)*t
Now we need to evaluate this in the value of t such that the side length is equal to 20 in:
S(t) = 20in = (2.29 in/min)*t
(20in)/(2.29 in/min) = t = 8.73 min
Then we will get:
A'(8.73min) = 2*(31.46 in^2/min^2)*8.73min = 439.29 in^2/min
This means that when the side length is equal to 20 inches, the surface area is increasing at a rate of 439.29 in^2 per minute.
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