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 89000 square centimeters?Note: The "altitude" is the "height" of the triangle in the formula "Area=(1/2)*base*height". Draw yourself a general "representative" triangle and label the base one variable and the altitude (height) another variable. Note that to solve this problem you don't need to know how big nor what shape the triangle really is.
A = .5 b h
dA/dt = .5 b dh/dt + .5 h db/dt
To solve this problem, we can use the formula for the area of a triangle:
Area = (1/2) * base * height
Given that the altitude (height) is increasing at a rate of 2500 cm/min and the area is increasing at a rate of 4000 cm^2/min, we can differentiate the formula with respect to time to relate the rates:
d(Area)/dt = (1/2) * base * d(height)/dt + (1/2) * height * d(base)/dt
Now, we plug in the given rates of change:
4000 = (1/2) * base * 2500 + (1/2) * 11500 * d(base)/dt
Since we know the values for the altitude (height) and the area, we can substitute them into the equation:
89000 = (1/2) * base * 2500 + (1/2) * 11500 * d(base)/dt
Now, we can solve for d(base)/dt, the rate at which the base is changing:
d(base)/dt = (2 * (89000 - (1/2) * base * 2500)) / (11500)
Simplifying further:
d(base)/dt = (178000 - 1250 * base) / 11500
Therefore, the rate at which the base of the triangle is changing when the altitude is 11500 cm and the area is 89000 cm^2 is given by the expression (178000 - 1250 * base) / 11500.