**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

**parameter**s 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:**Area of the rectangle**$A=49\ {\rm ft}^2$- The
**Length of the Rectangle**is assumed as $L$. - The
**Width of the Rectangle**is assumed as $W$. **Perimeter**$P$ is assumed to be**minimum**.

**Length**$L$ and

**Width**$W$ of a

**rectangle**that will result in the

**minimum Perimete**r $P$. As we know, the

**area of the rectangle**is defined as:\[A=L\times W\]\[49=L\times W\] 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 $32\ {\rm ft}^2$.

**Solution**Given that:

**Area of the rectangle**$A\ =\ 32\ \ {\rm ft}^2$**Perimeter**$P$ is assumed to be**minimum**.

**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**$A$:\[32\ {\rm ft}^2\ =\ 4\sqrt2\ ft\ \times\ W\]\[W\ =\ \frac{32\ {\mathrm{ft}}^2}{4\sqrt2\ ft\ }\]\[W\ =\ 4\sqrt2\ ft\]