Arectangular 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 12 cm, the edge length of the base is decreasing at a rate of 2 cm/min, the height of the box is 6 cm, and the height is increasing at a rate of 1 cm/min.

It is given in the question that

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

And the edge length of the base is 12 cm, the edge length of the base is decreasing at a rate of 2 cm/min, the height of the box is 6 cm, and the height is increasing at a rate of 1 cm/min.

Here we differentiate V with respect to t, and we use product rule, that is

Substituting the given values , we will get

" alt="dV/dt = 2(12)(6)(-2)+ 12^2(1)" />" />

So at that moment, the volume is decreasing at the rate of

the answer is ! ?

answer: bearing in mind that a rectangle has a perimeter of length + length + width + width.

step-by-step explanation:

p = perimeter

l = length

x=width

p = 2l + 2x

p = 2(l+x)     < ==== perimeter of the base of the box.