Question

A cube is contracting so that its surface area decreases at the constant rate of 72 in^2/sec. Determine how fast the Volume is changing at the instant when the surface area is 54 ft^2. (Barron's AP Calculus, 11th Ed., Bock & Hockett, copyright 2012, Barron's Educational Series, Inc., pg 496, problem 22). Book says dV/dt = -3/8. I get: dS/dt = dS/dx (dx/dt) = 72 in^2/sec = 1/2 ft^2/sec. Since S-Area = x^2, dS/dx = 2x. Then dx/dt = 1/(4x). Volume = x^3. dV/dt = (3x^2)(dx/dt) = (3/4)x . With x= sqr(54 ft^2)., dV/dt = (9/4)(sqr(6)) not = 3/8 ! (Ignoring negative sign). Is book wrong, or me?

Answer 1

Thinking of all 3 sides shrinking, or varying the same amount, I changed my equation to:
dV/dt = 3(x(dx/dt))^2(dx/dt)
dV/dt = 3x^2(dS/dt)(dx/dt)
dV/dt = (3x^2)(1/2)(1/(4x))
dV/dt = (3/8)x
Note the slope of the Volume is (3/8)
Is that what the question was asking, rather than the actual value of dV/dt at x = sqr(54)? Which would be
dV/dt = (3/8)sqr(54) = (9/8)sqr(6).

Answer 2

I ignored the negative sign again, which just means the Volume is decreasing (getting smaller as time goes on)