High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize High School Geometry Textbook (Common Core) > 2D Measurements

Optimal Area & Perimeter of a Rectangle

Area & PerimeterPrevious Section
Surface Area & Volume of a PrismNext Section
Popular Courses
Find My Course
0:00 / 0:00

Optimal Area & Perimeter of a Rectangle

Sometimes in geometry problems, we will be asked to determine the optimal / optimum measurement, this means the most desirable value for our measurement.

Examples
  • If we want to build a fence for our garden and want to save money on the fencing material, then the optimal perimeter is the
    smallest possible
    perimeter
  • if we want to build a fence for our garden using as much fencing material as possible, then the optimal perimeter is the
    largest possible
    perimeter
  • If we want to design a soda can that can hold the most amount of sode, then the optimum volume is the
    largest possible
    volume
  • If we want to construct a cardboard box and want to pay the least amount of money in materials, then the optimal surface area is the
    smallest possible
    surface area
0:00 / 0:00

Example: Determining Optimal Area Given Perimeter

A rectangle has a perimeter of 16 cm.

a) List all of the possible dimensions (length and width) that have integer values.

Since the perimeter of a rectangle is 2(l+w)2(l+w), we have the following possible dimensions:

lengthwidthPerimeter in cmArea in cm2172(1+7)=16 ✓1×7=7262(2+6)=16 ✓2×6=12352(3+5)=16 ✓3×5=15442(4+4)=16 ✓4×4=16532(5+3)=16 ✓5×3=15622(6+2)=16 ✓6×2=12712(7+1)=16 ✓7×1=7\begin{array}{|c|c|c|c|} \hline \text{length}&\text{width}&\text{Perimeter in cm}&\text{Area in cm}^2\\ \hline 1&7&2(1+7)=16 ~\checkmark&1\times7=7\\ \hline 2&6&2(2+6)=16~\checkmark&2\times6=12\\ \hline 3&5&2(3+5)=16~\checkmark&3\times5=15\\ \hline 4&4&2(4+4)=16~\checkmark&4\times4=16\\ \hline 5&3&2(5+3)=16~\checkmark&5\times3=15\\ \hline6&2&2(6+2)=16~\checkmark&6\times2=12\\ \hline7&1&2(7+1)=16~\checkmark&7\times1=7\\ \hline \end{array}

b) What is the largest area of such a rectangle?

The largest area is 16 cm216~\text{cm}^2, which occurs when the rectangle has the same length as the width -- meaning when it is a square!

c) If you were to guess, what would be the largest possible area of a rectangle that has a perimeter of 36 cm?

The largest possible area of this rectangle should be when the shape is a square with a perimeter of 36 cm.

Since a square has 4 equal side lengths, we can divide 36 by 4 to get that each side length is 9 cm9\text{ cm}.

Meaning that the area is 81 cm281~\text{cm}^2.
0:00 / 0:00

Example: Determining Optimal Perimeter Given Area

A rectangle has an area of 36 cm2.

a) List all of the possible dimensions (length and width) that have integer values.

Since the area of a rectangle is l×wl\times w, we have the following possible dimensions:

lengthwidthPerimeter in cmArea in cm21362(1+36)=741×36=36 ✓2182(2+18)=402×18=36 ✓3122(3+12)=303×12=36 ✓492(4+9)=264×9=36 ✓662(6+6)=246×6=36 ✓942(9+4)=269×4=36 ✓1232(12+3)=3012×3=36 ✓1822(18+2)=4018×2=36 ✓3612(36+1)=7436×1=36 ✓\begin{array}{|c|c|c|c|} \hline \text{length}&\text{width}&\text{Perimeter in cm}&\text{Area in cm}^2\\ \hline 1&36&2(1+36)=74 &1\times36=36~\checkmark\\ \hline 2&18&2(2+18)=40&2\times18=36~\checkmark\\ \hline 3&12&2(3+12)=30&3\times12=36~\checkmark\\ \hline 4&9&2(4+9)=26&4\times9=36~\checkmark\\ \hline 6&6&2(6+6)=24&6\times6=36~\checkmark\\ \hline 9&4&2(9+4)=26&9\times4=36~\checkmark\\ \hline 12&3&2(12+3)=30&12\times3=36~\checkmark\\ \hline 18&2&2(18+2)=40&18\times2=36~\checkmark\\ \hline 36&1&2(36+1)=74&36\times1=36~\checkmark\\ \hline \end{array}

b) What is the smallest perimeter of such a rectangle?

The smallest perimeter is 24 cm24~\text{cm}, which occurs when the rectangle has the same length as the width -- meaning when it is a square!

c) If you were to guess, what would be the smallest possible perimeter of a rectangle that has an area of 81 cm2?

The smallest possible perimeter of this rectangle should be when the shape is a square with an area of 81 cm2.

A square with an area of 81 cm2 will have a side length of 81=9 cm\sqrt{81}=9~\text{cm}.

Meaning that the perimeter if 4×9=36 cm4\times 9=36~\text{cm}.
0:00 / 0:00

Optimal Areas & Perimeters of Rectangles

Given a fixed perimeter PP, the rectangle with the largest possible area is a
square
.
The largest possible area is
A=(P4)2\boxed{A=\left(\dfrac{P}{4}\right)^2}.


Given a fixed area AA, the rectangle with the smallest possible perimeter is a
square
.
The smallest possoble perimeter is
P=4×A\boxed{P=4\times\sqrt{A}}

Practice: Optimal Area of a Rectangle

Determine the largest area of the rectangle with the following perimeters.

a) 144cm144 cm

b) 64ft64 ft

c) 45"45"

Practice: Optimal Perimeter of a Rectangle

Determine the smallest perimeter of the rectangle with the following areas.

a) 144 cm2144 ~cm^2

b) 64 ft264 ~ft^2

c) 45 inch245 ~inch^2

Practice: Optimal Area & Perimeter

Ella wants to build a new rectangular playground with an area of 1500 ft2. If the fencing material costs $5/ft, what is the cheapest she can build the fence around this playground for if she cannot buy the fencing material in a fraction of a foot (for example, she cannot buy 1.5 feet of fencing material, she would need to buy 2 feet of fencing material)?

Practice: Optimal Area & Perimeter

Angie is planning to build a new rectangular kitchen and is looking at the cost of baseboards, which is the material that is installed along the bottom of the walls to hide the gap between the floor and the wall. She will only need to install baseboards along 3 walls since the 4th wall will be covered with kitchen cabinets. What is the largest kitchen she can build if she only plans on buying 45 ft of baseboards? (There's no need to account for doors and other building requirements)