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\textbf{Applications of Partial Differentiation}

\textit{\textbf{Extremes within restricted domains}}
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\textbf{Question}

A building developer has bought a 10 hectare plot of land. On it he
could build 6 detached houses per hectare, 8 pairs of semi-detached houses per
hectare, or 12 flats per hectare.

The profits for each building are $\pounds$40,000 per detached house,
$\pounds$20,000 per pair of semi-detached houses and $\pounds$16,000
per flat.

However, council bylaws require that he build at least as many flats
as detached houses or pairs of semi-detached house.

How many of each building should he build to maximize his profit?


\textbf{Answer}

If the developer builds $x$ detached houses, $y$ pairs of
semi-detached houses and $z$ flats then his profit will be
$$P = 40000x+20000y+16000z.$$

The imposed constraints mean that
$$\frac{x}{6} + \frac{y}{8} + \frac{z}{12} \le 10 \Leftrightarrow
4x+3y+2z \le 240$$
and
$$z \ge x+y.$$

Obviously $x \ge 0$, $y \ge 0$ and $z \ge 0$.

Now, the planes $4x+3y+2z=240$ and $z=x+y$ intersect where
$6x+5y=240$. This means that the constraint region has the vertices
$(0,0,0)$, $(40,0,40)$, $(0,48,48)$ and $(0,0,120)$. These yield
revenues of $\pounds 0$, $\pounds 2240000$, $\pounds 1728000$ and
$\pounds 1920000$ respectively. So to maximize his profit, the
developer should build 40 detached house and 40 flats, but no pairs of
semi-detached houses.

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