A rectangular pig pen is built against an existing brick fence. $24$ m of fencing was used to enclose $70$ m$^2$. Find the dimensions of the pen.
If $2L+2W=24$ and $LW=70$
$2L+2W=24$ to $2L=24-2W$ to $L=12-W$
Substitute the answer into: $W(12-W)=70$ to $12W-W^2=70$ to $W^2-12W+70=0$
Quadratic formula with $a=1, b=-12$ and $c=70$ we get $\frac{12 \pm \sqrt{-136}}{2}$ and I cannot seem to go further than this. What can I do now?
If you have another way of solving this, I'd appreciate it if you'd share it with me too.
Ok I found a way to solve this.
$L+2W=24$ and $LW=70$
$L=24$-$2W$
Substitute into = $W(24$-$2W$)=$70$
$-2W^2$+$24W$-$70$=$0$
$W^2$-$12w$+$35$=$0$
W is 7 or 5.
So, $L$=$24-$2$\cdot $7
$L$=$10$
Or
$L$=$24-$2$\cdot $5
$L$=$14$
Answer: $7$ m by $10$ m or $5$ m and $14$ m.
We are told that a rectangular pig pen is built against a wall. That means three sides of the rectangle are bounded by a fence, while the fourth side of the pen is bounded by a wall.
In the diagram, the thick side represents the wall, while the thin sides represent the fence. We are told that $24~\text{m}$ of fencing are used. Hence, $$24~\text{m} = l + 2w \tag{1}$$ We are told that the area of the rectangle is $70~\text{m}^2$. Hence, $$70~\text{m}^2 = lw \tag{2}$$ Solve equation $1$ for $l$. Substitute the resulting expression in the equation for the area to obtain a quadratic equation in $w$. Solve it for $w$. Since $w$ represents the width of the pen, discard any solutions for $w$ that are not positive. Substitute your value(s) for $w$ into the equation you obtained for $l$ to find the corresponding value for $l$.