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:
- 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.
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:
- Area of the rectangle $A\ =\ 32\ \ {\rm ft}^2$
- 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 $A$:\[32\ {\rm ft}^2\ =\ 4\sqrt2\ ft\ \times\ W\]\[W\ =\ \frac{32\ {\mathrm{ft}}^2}{4\sqrt2\ ft\ }\]\[W\ =\ 4\sqrt2\ ft\]
