Question

A triangle's base is increasing at the rate of 3 inches/sec, its height at 2 inches/sec. At what rate is the area of the rectangle increasing when base is 10 inches and height is 14 inches

Answer 1

dA/dt = 31 in²/ sec

Explanation:

The area of the triangle is:

A(t) = (1/2)*b*h where b is the base and h the height

Differentiation on both sides of the equation we obtain

dA/dt = (1/2)* db/dt * h + (1/2)* dh/dt * b (1)

db/dt = 3 in/sec

dh/dt = 2 n/sec

b = 10 in

h = 14 in

By substitution in (1)

dA/dt = 0,5* 3*14 + 0,5* 2*10

dA/dt = 21 in² /sec + 10 in²/sc

dA/dt = 31 in²/ sec