im not sure how to do this

Question

The altitude of a triangle is increasing at a rate of 1.5 centimeters/minute while the area of the triangle is increasing at a rate of 1 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7.5 centimeters and the area is 99 square centimeters?

Answer 1

h'= 1.5

A'= 1

A= 99, h= 7.5

what is b'?

A=bh/2

2A =bh

A=99, b=7.5 solve for h

b= 2A/h = 2(99)/7.5= 26.4

take derivatives on both sides of 2A =bh and substitute values

2A' = bh' + hb'

solve h' -=(2A' - hb')/h = (2(1)(99) -26.4(1.5))/7.5

= (2-39.6)/7.5= -37.6/7.5 = about -5 cm per minute

h'= -5

altitude or height is decreasing about 5 centimeters per minute

base is increasing 50% faster than the Area is increasing.

which suggests altitude is decreasing

Area = 99 cm^2

base = 7.5 cm

A = bh/2

99 = 7.5h/2

198 = 7.5h

h = 198/7.5 = 26.4 cm

check on the answer

in one minute, Area increases from 99 to 100

in one minute base increases to 7.5+1.5 = 9

then the height increases from 26.4 to 47.4

100 = bh/2 = 9(47.4)2 =

if height decreases 5 cm per minute then it shrinks from 26.4 to 21.4

A=bh/2

100 = 9(21.4)/2 = 4.5(21.4) =96.3, close enough, especially as we rounded 5.01 down to 5