The altitude of a triangle is increasing at a rate of 2500 centimeters/minute while the area of the triangle is increasing at a rate of 4000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 11500 centimeters and the area is 84000 square centimeters?
Area = (1/2) bh , where b is the base and h is the height
d(Area)/dt = (1/2)b dh/dt + (1/2)h db/dt
when h = 11500 , A = 84000
84000 = (1/2)(b)(11500)
b = 14.609
in derivative equation ....
4000 = (1/2)(14.609)(2500) + (1/2)(11500)db/dt
db/dt = (8000 - 14.609(2500))/11500
= -2.48 cm/min
check my arithmetic, the negative seems to suggest that at that moment the base is decreasing.
Try considering the formula for the area of a triangle: A = (1/2)bh.
To find the rate at which the base of the triangle is changing, we need to use the formula for the area of a triangle. The area of a triangle is given by the equation:
Area = (1/2) * base * altitude
Differentiating both sides of the equation with respect to time (t), we get:
d(Area) / dt = (1/2) * d(base) / dt * altitude + (1/2) * base * d(altitude) / dt
Given that d(altitude) / dt = 2500 cm/min and d(Area) / dt = 4000 cm²/min, we can substitute these values into the equation:
4000 = (1/2) * d(base) / dt * 11500 + (1/2) * base * 2500
Now, we can solve for d(base) / dt:
d(base) / dt * 11500 = 8000 - 1250 * base
To find the value of base, we can use the equation for the area of the triangle and substitute the given values:
84000 = (1/2) * base * 11500
Simplifying, we find:
base = (2 * 84000) / 11500
Now we can substitute this value of base back into the equation:
d(base) / dt * 11500 = 8000 - 1250 * ((2 * 84000) / 11500)
Simplifying this equation gives us:
d(base) / dt = (8000 - 1250 * ((2 * 84000) / 11500)) / 11500
Now we can calculate this expression to find the rate at which the base of the triangle is changing.