the altitude of a triangle is increasing at a rate of 1 centimeters/minute while the area of the triangle is…

Explanation:

Step1: Recall the area formula of a triangle

The area formula of a triangle is (A=\frac{1}{2}bh), where (A) is the area, (b) is the base, and (h) is the altitude. Differentiate both sides with respect to time (t) using the product rule ((uv)^\prime = u^\prime v+uv^\prime). Here (u = b) and (v = h), so (\frac{dA}{dt}=\frac{1}{2}(b\frac{dh}{dt}+h\frac{db}{dt})).

Step2: Find the base when (A = 93) and (h=10.5)

From (A=\frac{1}{2}bh), we can solve for (b). Substitute (A = 93) and (h = 10.5) into (A=\frac{1}{2}bh). Then (93=\frac{1}{2}b\times10.5), so (b=\frac{93\times2}{10.5}=\frac{186}{10.5}=\frac{1860}{105}=\frac{124}{7}) cm.

Step3: Substitute the known values into the differentiated formula

We know that (\frac{dA}{dt}=3) (area increasing rate), (\frac{dh}{dt}=1) (altitude increasing rate), (h = 10.5), and (b=\frac{124}{7}). Substitute into (\frac{dA}{dt}=\frac{1}{2}(b\frac{dh}{dt}+h\frac{db}{dt})): (3=\frac{1}{2}(\frac{124}{7}\times1 + 10.5\times\frac{db}{dt})). First, multiply both sides by (2): (6=\frac{124}{7}+10.5\times\frac{db}{dt}). Then, (10.5\times\frac{db}{dt}=6-\frac{124}{7}=\frac{42 - 124}{7}=\frac{- 82}{7}). Since (10.5=\frac{21}{2}), we have (\frac{db}{dt}=\frac{-82}{7}\div\frac{21}{2}=\frac{- 82\times2}{7\times21}=\frac{-164}{147}\approx - 1.116) cm/min.

Answer:

(\frac{-164}{147}\approx - 1.12) cm/min