A rectangular box has a square base with an edge length of x cm and a height of h cm. The volume of the box is given by V = x2h cm3. Find the rate at which the volume of the box is changing when the edge length of the base is 4 cm, the edge length of the base is increasing at a rate of 2 cm/min, the height of the box is 15 cm, and the height is decreasing at a rate of 3 cm/min.

  192 cm^3/min

Step-by-step explanation:

Differentiating the volume expression, we get ...

  dV/dt = 2xh(dx/dt) +x^2(dh/dt)

We are given that ...

  x = 4 cm, dx/dt = 2 cm/min, h = 15 cm, dh/dt = -3 cm/min

Putting these values into the formula for volume rate of change, we get ...

  dV/dt = 2(4 cm)(15 cm)(2 cm/min) +(4 cm)^2(-3 cm/min)

  = 240 cm^3/min -48 cm^3/min

  dV/dt = 192 cm^3/min

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step-by-step explanation:

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

hope this : )