Hi lilkrazyrae,
This is a problem about related rates.
lilkrazyrae said:
The volume of a rectangular box with square base remains constant at 400cm^3 as the area of the base increases at a rate of 2 cm^2/s. Find the rate at which the height of the box is decreasing when each side of the base is 16 cm long.
What we have to do is find an equation for the volume, and then differentiate it with respect to time. After this, we will have enough information to solve for the rate at which the height is decreasing.
. . . .
From the diagram, a formula for the volume of the box is V=A*h (the volume of a rectangular box is the area of the base times the height). Furthermore, we know the volume of the box to be a constant 400 cm<sup>3</sup>. Thus, we obtain the formula:
. . . . A*h=400
Now, the problem asks for the
rate of change of the height. Therefore, you must involve derivatives in this equation somehow.
One way to do this is to differentiate both sides of the equation with respect to t. We have a constant on the right-hand side of the equation, which differentiates to 0. On the left-hand side, we must use the product rule (both A and h are functions of t, or in other words, they both change with time). That makes the forumla:
. . . . A'h + Ah' = 0
We are looking for the rate of change of the height and must therefore solve for h':
. . . .
All of the information on the right-hand side of this equation is given or can be found. We know A', the rate of change of the base area, to be 2 cm<sup>2</sup>/s. We also know that "each side of the base is 16 cm long." That makes the area, A, equal to 16*16=256 cm<sup>2</sup>. Finally, we can use the volume formula, V=A*h, to solve for h:
. . . .
Combine all this information together, and you can finally compute h':
. . . .
Thus, the height h is decreasing at a rate of
(25/2048) cm/s.
