# Chapter 2 - Section 2.8 - Modeling Using Variation - Exercise Set - Page 424: 26

a) $450$ houses can be served by a water pipe with a 30-centimeter diameter.   b) A water pipe with $25\text{ centimeter}$ diameter can serve a new subdivision of 1250 houses.

#### Work Step by Step

(a) We have, \begin{align} & h\propto {{d}^{2}} \\ & h=k{{d}^{2}} \\ \end{align} Here, h is the number of houses served and d is the diameter of the pipe. Substitute the value of d and h to get the value of k. \begin{align} & h=k{{d}^{2}} \\ & 50=k\cdot {{(10)}^{2}} \\ & k=\frac{50}{100} \\ & k=0.5 \end{align} Substitute $d=30$ and $k=0.5$in $h=k{{d}^{2}}$. \begin{align} & h=0.5{{d}^{2}} \\ & h=0.5{{(30)}^{2}} \\ & h=450 \\ \end{align} Therefore, 450 houses can be served by the water pipe with a 30-centimeter diameter.   (b) Substitute $h=50$ and $d=10$ in $h=k{{d}^{2}}$ \begin{align} & h=k{{d}^{2}} \\ & 50=k\cdot {{(10)}^{2}} \\ & k=\frac{50}{100} \\ & k=0.5 \end{align} As,$h=0.5{{d}^{2}}$ For $h=1250$ , \begin{align} & h=0.5{{d}^{2}} \\ & 1250=0.5{{d}^{2}} \\ & {{d}^{2}}=\frac{1250}{0.5} \\ & {{d}^{2}}=625 \\ & d=\sqrt{625} \\ & d=25 \\ \end{align} A water pipe with a 25-centimeter diameter can serve a new subdivision of 1250 houses.

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