The length of a rectangle is increasing at a rate of $7 \mathrm{~cm} / \mathrm{s}$ and its width is increasing at a rate of $6 \mathrm{~cm} / \mathrm{s}$. When the length is $9 \mathrm{~cm}$ and the width is $6 \mathrm{~cm}$, how fast is the area of the rectangle increasing (in $\mathrm{cm}^{2} / \mathrm{s}$ )? Enter an exact number. Need Help? Read It Watch it Master It

Step 1 ：The area of a rectangle is given by the formula \(A = lw\), where \(l\) is the length and \(w\) is the width. We are given that \(l\) is increasing at a rate of \(7 \mathrm{~cm} / \mathrm{s}\) and \(w\) is increasing at a rate of \(6 \mathrm{~cm} / \mathrm{s}\). We are asked to find how fast the area is increasing when \(l = 9 \mathrm{~cm}\) and \(w = 6 \mathrm{~cm}\). This is a problem of related rates.

Step 2 ：We can differentiate the area formula with respect to time \(t\) to get an equation that relates the rates of change of \(A\), \(l\), and \(w\).

Step 3 ：The derivative of the area with respect to time is given by \(\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}\).

Step 4 ：Substituting the given values, we get \(\frac{dA}{dt} = 9\cdot6 + 6\cdot7 = 96\).

Step 5 ：Final Answer: The rate at which the area of the rectangle is increasing is \(\boxed{96 \, \mathrm{cm}^{2}/\mathrm{s}}\).