5. The original 24 m edge length x of a cube decreases at the rate of 2 m/min.a. When x 2 m, at what rate does the cube's surface area change?b. When x 2 m, at what rate does the cube's volume change?m2/min.a. When x 2 m, the surface area is changing at a rate of(Type an integer or a decimal. Do not round.)m /min.b. When x 2 m, the volume is changing at a rate of(Type an integer or a decimal. Do not round.)

Question
Asked Oct 14, 2019

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5. The original 24 m edge length x of a cube decreases at the rate of 2 m/min.
a. When x 2 m, at what rate does the cube's surface area change?
b. When x 2 m, at what rate does the cube's volume change?
m2/min.
a. When x 2 m, the surface area is changing at a rate of
(Type an integer or a decimal. Do not round.)
m /min.
b. When x 2 m, the volume is changing at a rate of
(Type an integer or a decimal. Do not round.)
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5. The original 24 m edge length x of a cube decreases at the rate of 2 m/min. a. When x 2 m, at what rate does the cube's surface area change? b. When x 2 m, at what rate does the cube's volume change? m2/min. a. When x 2 m, the surface area is changing at a rate of (Type an integer or a decimal. Do not round.) m /min. b. When x 2 m, the volume is changing at a rate of (Type an integer or a decimal. Do not round.)

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Expert Answer

Step 1

According to the given information:

Length of the cube = ‘x’ m

Rate at which the length decreases (dx/dt) = -2m/min

Step 2

For part (a) it is needed to calculate the rate of ...

surface area of cube(S) 6x side2
=6x2
dS
= 12x
dt
dt
ds
(x 2m) 12(2) (-2)
dt
=-48m2 /min
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surface area of cube(S) 6x side2 =6x2 dS = 12x dt dt ds (x 2m) 12(2) (-2) dt =-48m2 /min

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Math

Calculus

Derivative