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Mashutka [201]
2 years ago
15

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm^2/

min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 120 cm^2?
Mathematics
1 answer:
VARVARA [1.3K]2 years ago
7 0

Answer:

The base is decreasing at 2 cm/min.

Step-by-step explanation:

The area (A) of a triangle is given by:

A = \frac{1}{2}bh   (1)

Where:

b: is the base

h: is the altitude = 10 cm

If we take the derivative of equation (1) as a function of time we have:

\frac{dA}{dt} = \frac{1}{2}(\frac{db}{dt}h + \frac{dh}{dt}b)

We can find the base by solving equation (1) for b:

b = \frac{2A}{h} = \frac{2*120 cm^{2}}{10 cm} = 24 cm

Now, having that dh/dt = 1 cm/min, dA/dt = 2 cm²/min we can find db/dt:

2 cm^{2}/min = \frac{1}{2}(\frac{db}{dt}*10 cm + 1 cm/min*24 cm)

\frac{db}{dt} = \frac{2*2 cm^{2}/min - 1 cm/min*24 cm}{10 cm} = -2 cm/min    

         

Therefore, the base is decreasing at 2 cm/min.

               

I hope it helps you!  

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