Application of quadratic functions

Question

Kevin wants to fence a rectangular garden using $14$ rails of $8$-foot rail, which cannot cut. What are the dimensions of the rectangle that will maximize the fenced area?

So the number of rails in each dimension the rectangle could be either $2$ in the width by $5$ in the length $($area $2(8)\cdot5(8)=640$ft$^2)$ or, $3$ in the width by $4$ in the length $($area $3(8)\cdot4(8)=1,152$ ft$^2)$, $1$ in the width and $6$ in the length $($area $1(8)\cdot6(8)=384$ft$^2)$.

Therefore the dimensions that maximize the fenced area are $24$ feet by $32$ feet. But I did that by trial and error. Not using a quadratic function.

This is what I did using quadratic function.

Let be $w$ the number of rails in the width and $l$ the number of rails in the length.

The perimeter would be:

$$2w+2l=14$$ $$\frac{2w}{2}+\frac{2l}{2}=\frac{14}{2}$$ $$w+1=7$$ $$w=7-l$$

The area would be:

$$A=w\cdot l$$ $$A=(7-l)\cdot l$$ $$A=7l-l^2$$

$$l=\frac{-7}{2(-1)}$$ $$l=\frac{-7}{-2}$$ $$l=\frac{7}{2}$$ $$l=3.5$$ $$l\approx 4$$

Shall I round this to $4~$? Since I can't cut the rail?

$$w=7-l$$ $$w=7-4$$ $$w=3$$

So the dimension of the length is $4(8)=32$ft and the width is $3(8)=24$ft

Is this right?

Answer 1

Yes, you are right.

Notice, a rectangle will have the maximum area only when its length and width become equal i.e. when it becomes a square. But since you can't cut a rail into pieces so you must have a rectangle which is the best approximation to a square for given conditions.

Therefore a rectangle of size $4\times 3$ is the best approximation to the square of size $3.5\times 3.5$ to have maximum area for given conditions in this case.

The red area shows the reduction ($=16\ \text{ft}^2$) in area of the rectangular garden of size $4\times 3$ which is best approximated to a square of size $3.5\times 3.5$ satisfying given conditions (as shown in figure).

Answer 2

The closest we can come to a square is a rectangle with $2$ sets of adjacent sides, each with a semi-perimeter of $\frac{14}{2}=7$ rails. The best we can do from here is to make each of the $2$ "corners" have $3$ and $4$ rails, respectively. We can multiply $3\times4\times64=768$ or $24\times32=768$ to get the "maximum" area.

Answer 3

you did it correctly.

if rails are not to be cut then you can make approximation of rails $$3.5\approx 4\ \ rails$$ this way, rectangular garden will have $4$ rails along each of two length and 3 rails along each of two widths. this gives you a rectangular garden of dimensions $$(4\times8)*(3\times 8)\ ft^2$$ area of final rectangular garden $$32*24=768\ ft^2$$