The height of a triangle is increasing at a rate of 3 cm/min while the area of the triangle is increasing at a rate of 11 square cm/min. At what rate is the base of the triangle changing when the height is 5 centimeters and the area is 10 square centimeters

Step-by-step explanation:

The Area of a triangle is expressed as:

A = 1/2 bh

b is the base of the triangle

h is the height of the triangle

Given

dh/dt = 3cm/min

dA/dt = 11cm²/min

height h = 5cm

Area = 10cm²

Get the base first

10 = 1/2b(5)

20 = 5b

b = 20/5

b = 4cm

Differentiating the area with respect to time

A(b, h) = 1/2 bh

According to product rule

dA/dt = 1/2{b dh/dt + h db/dt}

Substitute the given parameters and get db/dt

11 = 1/2{4(3)+5db/dt}

11 = 1/2{12+5 db/dt}

11 = 6 + 5/2 db/dt

11-6 = 5/2 db/dt

5 = 5/2 db/dt

1 = 1/2 db/dt

cross multiply

2 = db/dt

db/dt = 2cm/min

Hence the rate at which the base of the triangle is changing id 2cm/min

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