ONE EIGHTH CONTEST 2025
Ghana National College - 65 points
Tema Secondary School - 33 points
Effiduase SHCS - 28 points
ONE EIGHTH CONTEST 2025
Mankranso SHS - 43 points
St Paul's SHS, Denu - 23 points
Adonten SHS - 14 points
ONE EIGHTH CONTEST 2025
St. Thomas Aquinas SHS - 51 points
Tamale SHS - 37 points
O'Reilly SHS - 25 points
PROBLEM OF THE DAY ROUND 3
QUESTION
A rectangle of length $x$ centimeters and width $y$ centimeters has a constant area $96\text{ cm}^2$. Express the perimeter $P$ of the rectangle in terms of $x$. Hence, find the least possible value of the perimeter.
ANSWER: $P = 2x + \dfrac{192}{x}$, Least Perimeter $= 16\sqrt{6}\text{ cm}$
SOLUTION
$\text{Area} = x \times y$
$\text{Area} = 96$
$y = \dfrac{96}{x}$
$P = 2(x + y)$
$P = 2\left(x + \dfrac{96}{x}\right)$
$P = 2x + \dfrac{192}{x}$
$\dfrac{dP}{dx} = 2 - \dfrac{192}{x^2}$
Equate to zero:
$2 - \dfrac{192}{x^2} = 0$
$2 = \dfrac{192}{x^2}$
$2x^2 = 192$
$x^2 = 96$
$x = 4\sqrt{6}\text{ cm}$
Check for minimum:
$\dfrac{d^2P}{dx^2} = \dfrac{384}{x^3}$
$\dfrac{384}{(4\sqrt{6})^3} > 0 \implies \text{Minimum}$
Substitute $x = 4\sqrt{6}$:
$P = 2(4\sqrt{6}) + \dfrac{192}{4\sqrt{6}}$
$P = 8\sqrt{6} + \dfrac{48}{\sqrt{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}}$
$P = 8\sqrt{6} + 8\sqrt{6}$
$P = 16\sqrt{6}\text{ cm}$
PRACTICE QUESTIONS
1. Find the least perimeter of a rectangle of area 81 cm².
ANSWER: $36$ cm
SOLUTION 1
For a rectangle of constant area $A$, the perimeter is least when it is a square:
$P_{\min}=4\sqrt{A}$
$P_{\min}=4\sqrt{81}$
$P_{\min}=4(9)$
$P_{\min}=36$
SOLUTION 2
Let the length be $x$, so the width is $\dfrac{81}{x}$.
$P=2x+\dfrac{162}{x}$
$\dfrac{dP}{dx}=2-\dfrac{162}{x^2}$
$2-\dfrac{162}{x^2}=0$
$x^2=81$
$x=9$
$P=2(9)+\dfrac{162}{9}$
$P=18+18$
$P=36$
2. Find the least perimeter of a rectangle of area 100 cm².
ANSWER: $40$ cm
SOLUTION 1
For a rectangle of constant area $A$, the perimeter is least when it is a square:
$P_{\min}=4\sqrt{A}$
$P_{\min}=4\sqrt{100}$
$P_{\min}=4(10)$
$P_{\min}=40$
SOLUTION 2
Let the length be $x$, so the width is $\dfrac{100}{x}$.
$P=2x+\dfrac{200}{x}$
$\dfrac{dP}{dx}=2-\dfrac{200}{x^2}$
$2-\dfrac{200}{x^2}=0$
$x^2=100$
$x=10$
$P=2(10)+\dfrac{200}{10}$
$P=20+20$
$P=40$
3. Find the least perimeter of a rectangle of area 144 cm².
ANSWER: $48$ cm
SOLUTION 1
For a rectangle of constant area $A$, the perimeter is least when it is a square:
$P_{\min}=4\sqrt{A}$
$P_{\min}=4\sqrt{144}$
$P_{\min}=4(12)$
$P_{\min}=48$
SOLUTION 2
Let the length be $x$, so the width is $\dfrac{144}{x}$.
$P=2x+\dfrac{288}{x}$
$\dfrac{dP}{dx}=2-\dfrac{288}{x^2}$
$2-\dfrac{288}{x^2}=0$
$x^2=144$
$x=12$
$P=2(12)+\dfrac{288}{12}$
$P=24+24$
$P=48$
4. Find the least perimeter of a rectangle of area 49 cm².
ANSWER: $28$ cm
SOLUTION 1
For a rectangle of constant area $A$, the perimeter is least when it is a square:
$P_{\min}=4\sqrt{A}$
$P_{\min}=4\sqrt{49}$
$P_{\min}=4(7)$
$P_{\min}=28$
SOLUTION 2
Let the length be $x$, so the width is $\dfrac{49}{x}$.
$P=2x+\dfrac{98}{x}$
$\dfrac{dP}{dx}=2-\dfrac{98}{x^2}$
$2-\dfrac{98}{x^2}=0$
$x^2=49$
$x=7$
$P=2(7)+\dfrac{98}{7}$
$P=14+14$
$P=28$
5. Find the least perimeter of a rectangle of area 169 cm².
ANSWER: $52$ cm
SOLUTION 1
For a rectangle of constant area $A$, the perimeter is least when it is a square:
$P_{\min}=4\sqrt{A}$
$P_{\min}=4\sqrt{169}$
$P_{\min}=4(13)$
$P_{\min}=52$
SOLUTION 2
Let the length be $x$, so the width is $\dfrac{169}{x}$.
$P=2x+\dfrac{338}{x}$
$\dfrac{dP}{dx}=2-\dfrac{338}{x^2}$
$2-\dfrac{338}{x^2}=0$
$x^2=169$
$x=13$
$P=2(13)+\dfrac{338}{13}$
$P=26+26$
$P=52$
6. Find the least perimeter of a rectangle of area 225 cm².
ANSWER: $60$ cm
SOLUTION 1
For a rectangle of constant area $A$, the perimeter is least when it is a square:
$P_{\min}=4\sqrt{A}$
$P_{\min}=4\sqrt{225}$
$P_{\min}=4(15)$
$P_{\min}=60$
SOLUTION 2
Let the length be $x$, so the width is $\dfrac{225}{x}$.
$P=2x+\dfrac{450}{x}$
$\dfrac{dP}{dx}=2-\dfrac{450}{x^2}$
$2-\dfrac{450}{x^2}=0$
$x^2=225$
$x=15$
$P=2(15)+\dfrac{450}{15}$
$P=30+30$
$P=60$