4. Let x be the length of one of the equal sides of an isosceles triangle, and let
q be the angle between them. If x is increasing at the rate 1/12 m/hr, and q is increasing at the rate of p /180 radians/hr, then at what rate, in m2/hr, is the area of the triangle increasing when x = 12 m and q = p /4?
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A. 36(21/2) |
B. (73/2)(21/2) |
C. (31/2/2)+( p /5) |
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D. 21/2((1/2)+(1/5) p ) |
E. 21/2(6+(1/5) p ) |
|
Solution: The answer is D
Consider the following isosceles triangle;

Where h is the height of the triangle, b is the base, and
q is the angle between the equal lengths, X. Since the given triangle is isosceles we can define b and h as follows,
So that the area, A, is defined as,

Differentiating both sides of the above equation with respect to time t yields,

Since, dx/dt = 1/12 and d
q /dt = p /180. Then by substitution,
Finally, when x = 12 and
q = p /4,