The length of a rectangle is increasing at a rate of 6 cm/s and its width is increasing at a rate of 4 cm/s. when the length is 11 cm and the width is 5 cm, how fast is the area of the rectangle increasing?

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Mathematics

Gavin Hampton

28 September, 10:04

The length of a rectangle is increasing at a rate of 6 cm/s and its width is increasing at a rate of 4 cm/s. when the length is 11 cm and the width is 5 cm, how fast is the area of the rectangle increasing?

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Yadiel Bautista · Answer 1

This is a nice "rates of change" problem from Calculus.

Let the length and width of the rect. be L and W. We are given the following info:

dL/dt = 6 cm/s; dW/dt = 4 cm/s; L = 11 cm and W = 5 cm.

The area of the rect. is A = L*W. Differentiating,

dA/dt = L (dW/dt) + W (dL/dt).

Subst. the given info: dA/dt = (11 cm) (4 cm/sec) + (5 cm) (6 cm/sec).

Just evaluate this to find dA/dt: (44 + 30) cm^2/sec = 76 cm^2/sec