The area of the waterbed formed with a fixed length of fencing, has an
area given by a quadratic function.
- a) The expression is; 2·x + 2·y = 120
- d) A = x × y = x × (60 - x) = 60·x - x²
- e) Please find the required graph created with MS Excel, showing the maximum area.
- f) The maximum area of 900 cm² occurs at the point where x = 30 cm and y = 30 cm
Reasons:
The length of fencing available = 120 m
Length of the rectangular flowerbed = x cm
Width of the flowerbed = y cm
a) The perimeter of the flowerbed = The length of the fencing available
The perimeter of a rectangle = 2 × Length + 2 × Width
Therefore;
Perimeter of the flowerbed, P = 2·x + 2·y = 120
b) The equation, P = 2·x + 2·y = 120, can be rearranged as follows;
2·x + 2·y = 120
Therefore;
(2·x + 2·y) ÷ 2 = 120 ÷ 2 = 60
x + y = 60
Which gives;
y = 60 - x
c) The area of a rectangle = Length × Width
Therefore;
The area of the rectangular flowerbed is, A = x × y
b) From part (b) above;
y = 60 - x
From part (c);
A = x × y
Therefore;
A = x × (60 - x) = 60·x - x²
A = 60·x - x²
e) Please find attached the graph of the equation A = 60·x - x², created with MS Excel
From the graph, we have;
The maximum area (point) is 900 cm² when the width is x = 30 cm
f) The function for the area is a quadratic function (f(x) = a·x² + b·x + c)
The value of x at the maximum area is given as follows;
[tex]\displaystyle x = \mathbf{ -\frac{b}{2 \cdot a}}[/tex]
Where, A = 60·x - x², we have;
[tex]At \ the \ maximum \ area, \ \displaystyle x = -\frac{60}{2 \times (-1)} = 30[/tex]
Therefore;
Maximum area, [tex]A_{max}[/tex] = 60 × 30 - 30² = 900
The maximum area is 900 cm², at x = 30
y = 60 - x
Therefore;
y = 60 - 30 = 30
At the maximum area, x = 30, and y = 30
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