Area=length times width
area=10
length=2+√2
width=w
10=(2+√2)(w)
divide both sides by (2+√2)

if we were to rationalize the dnomenator
multiply by



the width is 10-5√2 m
<h3><u>Solution</u></h3>
<u>Given </u><u>:</u><u>-</u>
- Perimeter of rectangle = 72 cm
- The length is 3 more than twice the width.
<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>:</u><u>-</u>
<h3 /><h3>
<u>Explantion</u></h3>
<u>Using </u><u>Formula</u>

<u>Let,</u>
- Length of Rectangle = x cm
- Breadth of Rectangle = y cm
<u>According</u><u> to</u><u> question</u><u>,</u>
==> perimeter of Rectangle = 72
==> 2(x+y) = 72
==> x + y = 72/2
==> x + y = 36_________________(1)
<u>Again,</u>
==> x = 2y + 3
==> x - 2y = 3__________________(2)
<u>Subtract</u><u> </u><u>equ(</u><u>1</u><u>)</u><u> </u><u>&</u><u> </u><u>equ(</u><u>2</u><u>)</u>
==> y + 2y = 36 - 3
==> 3y = 33
==> y = 33/3
==> y = 11
<u>keep </u><u>in </u><u>equ(</u><u>1</u><u>)</u>
==> x - 2×11 = 3
==> x = 3 + 22
==> x = 25
<h3><u>Hence</u></h3>
- <u>Length</u><u> of</u><u> </u><u>Rectangle</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u> </u><u>cm</u>
- <u>Width </u><u>of </u><u>Rectangle</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u> </u><u>cm</u>
<h3>
<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u></h3>
<h3 />
A.) P(t) = 130t - 0.4t^4 + 1200
The population is maximum when P'(t) = 0
P'(t) = 130 - 1.6t^3 = 0
1.6t^3 = 130
t^3 = 81.25
t = ∛81.25 = 4.3 months.
Maximum population P(t)max = 130(4.3) - 0.4(4.3)^4 + 1200 = 1,622
b.) The rabbit population will disappear when P(t) = 0
P(t) = 130t - 0.4t^4 + 1200 = 0
t ≈ 8.7 months