1. ## Optimization Problem

A builder must completely fence in 3 adjacent rectangular lots along a roadway. Each lot must have an area of 675 square metres. Fencing along the roadway costs $25/metre while side and back fencing costs$10/metre. Find the dimensions of each lot that minimize the total cost of fencing.

2. Hello, NAPA55!

The answers come out rather ugly . . .

A builder must completely fence in 3 adjacent rectangular lots along a roadway.
Each lot must have an area of 675 square metres.
Fencing along the roadway costs $25/metre while side and back fencing costs$10/metre.
Find the dimensions of each lot that minimize the total cost of fencing.
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x         x         x

There are $3x$ m of fencing along the road.
At $25/m, its cost is: . $(25)(3x) \:= \:75x$ dollars. There are $3x$ m of back fencing and $4y$ m of side fencing. At$10/m, its cost is: . $10(3x + 4y) \:=\:30x + 40y$ dollars.

The total cost is: . $C \;=\;75x + (30x + 40y) \;=\;105x + 40y$ dollars. .[1]

Each lot has an area of 675 mē: . $xy \:=\:675\quad\Rightarrow\quad y \:=\:\frac{675}{x}$

Substitute into [1]: . $C \;=\;105x + 40\left(\frac{675}{x}\right) \quad\Rightarrow\quad C \;=\;105x + 25,000x^{-1}$

And that is the function we must minimize . . .

3. Just to finish off what Soroban started...

Note: Remember your domain $C =105x + 25,000x^{-1}\;\;\;\;0< x <\infty$

$\frac {dC}{dx}=105-25000x^{-2}$, set the derivative equal to zero...

$25000x^{-2}=105\to x^2=\frac{5000}{21}\to x=50\sqrt{\frac{2}{21}}$. (x cannot be negative, as denoted by our domain, therefore we take the positive).

Sub x back in and you should get $y=\frac{27}{2}\sqrt{\frac{21}{2}}$

Therefore the dimension of each lot must be $50\sqrt{\frac{2}{21}}$ x $\frac{27}{2}\sqrt{\frac{21}{2}}$to minimize the cost of fencing.