The volume of a rectangular box with a square base remains constant at 500 cm3 as the area of the base increases at a rate of 6 cm2/sec. Find the rate at which the height of the box is decreasing when each side of the base is 15 cm long.

the rate at which the height of the box is decreasing is -0.0593 cm/s

Step-by-step explanation:

Given the data in the question;

Constant Volume of a rectangular box with a square base = 500 cm³

area of the base increases at a rate of 6 cm²/sec

so change in the area of the base with respect to time dA/dt = 6 cm²/sec

each side of the base is 15 cm long

so Area of the base = 15 cm × 15 cm = 225 cm²

the rate at which the height of the box is decreasing = ?

Now,

V = Ah

dv/dt = 0 ⇒ Adh/dt + hdA/DT = 0

⇒ dh/dt = -hdA/dt / A

we substitute

dh/dt = [ -( 500 / 225 ) × 6 ] / 225

dh/dt = [ -(2.22222 × 6)  ] / 225

dh/dt = [ -13.3333 ] / 225

dh/dt = -0.0593 cm/s

Therefore, the rate at which the height of the box is decreasing is -0.0593 cm/s

v v cj van config k tbs eng r.c r.c fcb fb gb frfr

-2

7 - 9 = -2

reversed:

-2 + 9 = 7