Question

The altitude of a triangle is increasing at a rate of 1 centimeters/minute while the area of the triangle is increasing at a rate of 4.5 square imeters / m * inute . At what rate is the base of the triangle changing when the altitude is 10 centimeters and the area is 95 square centimeters?

Answer 1

Solution:

The area of a triangle is expressed as

$\begin{gathered} A=\frac{1}{2}bh\text{ ----- equation 1} \\ where \\ A\Rightarrow area \\ b\Rightarrow base \\ h\Rightarrow altitude \end{gathered}$

By taking the derivative of equation 1 with respect to time t, we have

$\frac{dA}{dt}=\frac{1}{2}b\frac{dh}{dt}+\frac{1}{2}h\frac{db}{dt}\text{ ---- equation 2}$

Given that the altitude of the triangle is increasing at a rate of 1 cm/minute while the area of the triangle is increasing at a rate of 4.5 square cm/minute, this implies that

$\begin{gathered} \frac{dh}{dt}=\text{ 1 cm/minute} \\ \frac{dA}{dt}=4.5\text{ cm}^2\text{ per minute} \end{gathered}$

Given that the altitude and area are 10 centimeters and 95 square centimeters, this implies that

$\begin{gathered} h=10\text{ cm} \\ A=\text{ 95 cm}^2 \end{gathered}$

We can evaluate the base of the triangle from equation 1 as

$\begin{gathered} 95=\frac{1}{2}\times b\times10 \\ \Rightarrow5b=95 \\ divide\text{ both sides by 5} \\ \frac{5b}{5}=\frac{95}{5} \\ \Rightarrow b=19\text{ cm} \end{gathered}$

By substituting these values into equation 2, we have

$\begin{gathered} 4.5=\frac{1}{2}(19)(1)+\frac{1}{2}(10)(\frac{db}{dt}) \\ \Rightarrow4.5=9.5+5\frac{db}{dt} \\ add\text{ -9.5 to both sides} \\ -9.5+4.5=-9.5+9.5+5\frac{db}{dt} \\ \Rightarrow-5=5\frac{db}{dt} \\ thus, \\ \frac{db}{dt}=-1\text{ cm per minute} \end{gathered}$

Hence, the base is changing at the rate of

$\frac{db}{dt}=-1\text{ centimeters per minute}$