Solution:
The area of a triangle is expressed as
[tex]\begin{gathered} A=\frac{1}{2}bh\text{ ---- equation 1} \\ where \\ A\Rightarrow area \\ b\Rightarrow base \\ h\Rightarrow altitude \end{gathered}[/tex]
From equation 1, taking the derivative of A with respect to t, we have
[tex]\frac{dA}{dt}=\frac{1}{2}(b\frac{dh}{dt}+h\frac{db}{dt})----\text{ equation 2}[/tex]
Given that the altitude is increasing at a rate of 2.5 cm/minute while the area is increasing at a rate of 5 square cm /minute, this implies that
[tex]5=\frac{1}{2}(b(2.5)+h(\frac{db}{dt}))\text{ ---- equation 3}[/tex]
To evaluate the rate at which the base is changing when the altitude is 8.5 cm and the area is 84 square cm, we have
[tex]\begin{gathered} A=\frac{1}{2}bh \\ \Rightarrow84=\frac{1}{2}\times b\times8.5 \\ \Rightarrow b=19.76470 \end{gathered}[/tex]
By substitution, we have
[tex]\begin{gathered} 5=\frac{1}{2}(19.76470(2.5)+8.5(\frac{db}{dt})) \\ \Rightarrow\frac{db}{dt}=-4.63667 \end{gathered}[/tex]
Hence, the base changes at
[tex]4.63667\text{ cm/min}[/tex]
