Question 1 of 3
A developer wishes to construct a duplex (two residential units that are side-by-side, sharing
a common wall). The interior floors are to be congruent rectangles with areas of 1200 square
feet. The outside wall and the common wall that separates the residences are constructed
from cinderblocks. Let I represent the length of one residential unit of the duplex and w
represent its width.
Choose a function, in terms of I, that could be used to determine the total length of
cinderblocks needed for all walls (including both the exterior and common walls).

C(L) = 4L + 3600/L (assuming common wall is width)

C(L) = 3L + 4800/L (assuming common wall is length)

Step-by-step explanation:

The area of residence is 1200 sq feet.

LW = 1200

W = 1200/L

The length of one residence is L, and the width is 1200/L.

I assume the common wall is the width.

The total length of the duplex is 2L.

The width is W = 1200/L

Also, the perimeter of the entire duplex is 2L + 2L + W + W.

In addition, there is a common wall whose length is the width of the duplex.

The entire length of cinder block walls is 2L + 2L + W + W + W, but we know that W = 1200/L, so we get

C(L) = 4L + 3W

C(L) = 4L + 3(1200/L)

C(L) = 4L + 3600/L

This is with the assumption that the common wall is a width, not a length.

If the common wall is a length, then we get this.

The perimeter of the entire duplex is L + L + 2W + 2W.

The common wall is a length of the duplex.

The entire length of cinder block walls is L + L + 2W + 2W + L, but we know that W = 1200/L, so we get

C(L) = 3L + 4W

C(L) = 3L + 4(1200/L)

C(L) = 3L + 4800/L

l.jnl.j/n.m,lm,

step-by-step explanation:

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