# Find the length and width of a rectangle that has the given area and a minimum perimeter.

Area of Rectangle A = 49 ft^(2)The aim of this article is to find the dimensions of a rectangle whose area A is known and it is assumed that it has the minimum perimeter P.The basic concept behind this article is the Optimization Process to calculate the minimum possible values and the Area or Perimeter of a geometrical figure.The Optimization Process is utilized to calculate the values that represent the variable parameters of a given function. These values are then used to find the minimum points of the function.If we are given one fixed value for a function having $2$ variables $x$ and $y$, we use it to define both the parameters in terms of a single variable. After which the Optimization Process of Minimum function is performed by the process of differentiation of the given minimum function with respect to one variable.

## Expert Answer

Given that:
1. Area of the rectangle $A=49\ {\rm ft}^2$
2. The Length of the Rectangle is assumed as $L$.
3. The Width of the Rectangle is assumed as $W$.
4. Perimeter $P$ is assumed to be minimum.

Width of rectangle

We have to find the dimensions i.e., Length $L$ and Width $W$ of a rectangle that will result in the minimum Perimeter $P$.

Parameter of rectangle

As we know, the area of the rectangle is defined as:$A=L\times W$$49=L\times W$

Area of rectangle

As we are given a fixed value of Area $A$, we will use it to define both length $L$ and width $W$ in terms of a single parameter as follows:$W=\frac{49}{L}$The Perimeter $P$ of a rectangle is:$P=2L+2W$Substituting the value of $W$ in the above equation:$P=2L+2\left(\frac{49}{L}\right)$$P=2L+98L^{-1}$Now we will use the Optimization Process by differentiation of the given minimum function of Perimeter $P$ with respect to one variable i.e., Length $L$.$\frac{d}{dL}(P)=\frac{d}{dL}\left(2L+98L^{-1}\right)$$\frac{dP}{dL}=\frac{d}{dL}(2L)+\frac{d}{dL}(98L^{-1})$$0=2+(-98L^{-2})$$2-\frac{98}{L^2}=0$$L^2=49$$L=7ft$Substituting the value of $L$ in expression for Area $A$:$A=L\times W$$49{\rm ft}^2=7ft\times W$$W=\frac{49{\rm ft}^2}{7ft\ }$$W=7ft$

## Numerical Result

A rectangle with a minimum perimeter and fixed area of $49{\rm ft}^2$ has a length of $7ft$ and width of $7ft$.

## Example

Find the dimensions of a rectangle having a minimum perimeter and a fixed area of $32\ {\rm ft}^2$.SolutionGiven that:
1. Area of the rectangle $A\ =\ 32\ \ {\rm ft}^2$
2. Perimeter $P$ is assumed to be minimum.
As we know the area of a rectangle is defined as:$A\ =\ L\ \times\ W$$32\ =\ L\ \times\ W$$W\ =\ \frac{32}{L}$As we know that the Perimeter $P$ of a rectangle is expressed as follows:$P\ =\ 2L\ +2W$Substituting the value of $W$:$P\ =\ 2L\ +2\left(\frac{32}{L}\right)$$P\ =\ 2L\ +64L^{-1}$Now we will use the Optimization Process by differentiation.$\frac{d}{dL}\ (P)\ =\ \frac{d}{dL}\ \left(2L\ +64L^{-1}\right)$$0\ =\ 2\ +\ (-64L^{-2})$$2\ -\ \frac{64}{L^2}\ =\ 0$$L^2\ =\ 32$$L\ =\ 4\sqrt2\ ft$Substituting the value of $L$ in expression for Area $A$:$32\ {\rm ft}^2\ =\ 4\sqrt2\ ft\ \times\ W$$W\ =\ \frac{32\ {\mathrm{ft}}^2}{4\sqrt2\ ft\ }$$W\ =\ 4\sqrt2\ ft$