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

The area of a triangle is expressed as

`\begin{gathered} A=(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

`(dA)/(dt)=(1)/(2)b(dh)/(dt)+(1)/(2)h(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} (dh)/(dt)=\text{ 1 cm/minute} \\ (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=(1)/(2)* b*10 \\ \Rightarrow5b=95 \\ divide\text{ both sides by 5} \\ (5b)/(5)=(95)/(5) \\ \Rightarrow b=19\text{ cm} \end{gathered}`

By substituting these values into equation 2, we have

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

Hence, the base is changing at the rate of

`(db)/(dt)=-1\text{ centimeters per minute}`