he altitude of a triangle is increasing at a rate of 3.000 centimeters/minute while the area of the triangle is increasing at a rate of 4.000 square centimeters/minute. at what rate is the base of the triangle changing when the altitude is 9.500 centimeters and the area is 88.000 square centimeters?

A line segment passing through a triangle's vertex and running perpendicular to the line containing the base is the triangle's height in geometry. The extended base of the altitude is the name given to this line that contains the opposing side. The foot of the altitude exists the point at where the extended base and the height converge.

A triangle's altitude is determined by drawing a perpendicular line from its vertex to its opposite side. The altitude, sometimes referred to as the triangle's height, forms a right-angle triangle with the base.

Let A be the area of the triangle

h be the altitude

b be the base

The formula that relates the three is the area of a triangle

A = (1/2) bh

Note that anything listed as a "rate" is a derivative value.

dh (rate of change of altitude) = 3.000 cm/min

dA (rate of change of area) = 4 cm²/min

h = 9.500 cm

A = 88.000 cm²

Consider that both b and h are changing in this situation, so we have to treat them both as variables. This means we have to use the product rule, inserting the derivatives where applicable.

dA = (1/2) × db × h + (1/2)b × dh

substitute the values in the above equation, we get

4 = (1/2) × db × 9.5 + (1/2) b × 3.000

88 = (1/2) × b × 9.5

b = 18.52631

Now, looking back at our derivative

4 = (1/2) × db × 9.5 + (1/2) × 18.52631 × 3.000

simplifying the above equation, we get

4 = 4.75 db + 13.26316

b =[tex]$ -\frac{9.26316}{4.75d}[/tex]

To learn more about altitude of a triangle refer to:

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