The altitude of a triangle is increasing at a rate of 1.5 cm/min while the area of the triangle is increasing at a rate of 5 square cm/min. At what rate is the base of the triangle changing when the altitude is 9 cm and the area is 81 square cm?
Accepted Solution
A:
Answer: -1 8/9 cm/minStep-by-step explanation:The area of a triangle is given by ... A = (1/2)bhThen when the area is 81 cm² and the altitude is 9 cm, the base is ... 81 cm² = (1/2)(b)(9 cm) (81 cm²)/(4.5 cm) = b = 18 cmDifferentiating the area formula implicitly, you have ... A' = (1/2)(b'h +bh')Fill in the given values and solve for b': 5 cm²/min = (1/2)(b'(9 cm) +(18 cm)(1.5 cm/min)) 5 cm²/min = b'(4.5 cm) +13.5 cm²/min . . . . simplifySubtract 13.5 and divide by the coefficient of b': (-8.5 cm²/min)/(4.5 cm) = b' = -1 8/9 cm/minThe base is changing at the rate of -1 8/9 ≈ -1.889 cm per minute.