The altitude (i.e., height) of a triangle is increasing at a rate of 2.5 cm/minute while the area of the triangle is increasing at a rate of

Question

The altitude (i.e., height) of a triangle is increasing at a rate of 2.5 cm/minute while the area of the triangle is increasing at a rate of 2.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 7 centimeters and the area is 84 square centimeters?

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Allison 1 week 2021-10-09T20:28:56+00:00 1 Answer 0

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    2021-10-09T20:30:17+00:00

    Answer:

    The base is decreasing at a rate of 7.8571 centimeters per minute.

    Step-by-step explanation:

    We are given the following information in the question:

    The height of a triangle is increasing at a rate of 2.5 cm/minute

    \displaystyle\frac{dh}{dt} = 2.5 \text{ cm per minute}

    The area of the triangle is increasing at a rate of 2.5 square cm/minute.

    \displaystyle\frac{dA}{dt} = 2.5 \text{ square cm per minute}

    Area of triangle is given by:

    A = \displaystyle\frac{1}{2}\times b\times h

    where A is the area of triangle, b is the base of triangle and h is the height of the triangle.

    Differentiating, we get,

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

    We have to find rate of change of base of the triangle when the altitude is 7 centimeters and the area is 84 square centimeters

    h = 7 cm

    A  = 84 square centimeters

    84 = \displaystyle\frac{1}{2}\times 7\times b\\\\b = 24\text{ cm}

    Putting the values, we get:

    \displaystyle\frac{dA}{dt} = \frac{1}{2}\Bigg(b\frac{dh}{dt} + h\frac{db}{dt}\Bigg)\\\\2.5 = \frac{1}{2}\Bigg((24)(2.5) + (7)\frac{db}{dt}\Bigg)\\\\5 - 60 = 7\frac{db}{dt}\\\\\frac{dh}{dt} =-7.8571\text{ cm per minute}

    Thus, the base is decreasing at a rate of 7.8571 centimeters per minute.

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