Surface Area Rate F8Afb4

1. **State the problem:** We have a rectangular box with height $b$, length $b+4$, and width $b+1$. We want to find how fast the total surface area increases as $b$ increases. 2. **Formula for surface area of a rectangular box:** $$S = 2(lw + lh + wh)$$ where $l$ is length, $w$ is width, and $h$ is height. 3. **Substitute given dimensions:** $$l = b+4, \quad w = b+1, \quad h = b$$ 4. **Write surface area in terms of $b$:** $$S = 2\big((b+4)(b+1) + (b+4)b + (b+1)b\big)$$ 5. **Expand each product:** $$(b+4)(b+1) = b^2 + b + 4b + 4 = b^2 + 5b + 4$$ $$(b+4)b = b^2 + 4b$$ $$(b+1)b = b^2 + b$$ 6. **Sum inside parentheses:** $$b^2 + 5b + 4 + b^2 + 4b + b^2 + b = 3b^2 + 10b + 4$$ 7. **Surface area formula simplified:** $$S = 2(3b^2 + 10b + 4) = 6b^2 + 20b + 8$$ 8. **Find rate of change of surface area with respect to $b$:** $$\frac{dS}{db} = \frac{d}{db}(6b^2 + 20b + 8) = 12b + 20$$ 9. **Interpretation:** The rate at which the surface area increases as $b$ increases is given by $$\frac{dS}{db} = 12b + 20$$. This means for any value of $b$, the surface area increases at a rate of $12b + 20$ units squared per unit increase in $b$.