Question

Qsn 1.4.1 have never done a problem of this kind before so will need background explanation for every step in the answer.

Answer 1

Maximum area = 4900 m²

Explanation

1.4.1 The perimeter of a rectangle is calculated as follows:

`P=2(l+w)`

where l is the length and w is the width of the rectangle.

Substituting P = 280 m, l = 2x meters, and w = y meters, and solving for y:

`\begin{gathered} 280=2(2x+y) \\ (280)/(2)=(2(2x+y))/(2) \\ 140=2x+y \\ 140-2x=2x+y-2x \\ 140-2x=y \end{gathered}`

The area of a rectangle is calculated as follows:

`A=l\cdot w`

Substituting l = 2x and w = y = 140 - 2x:

`\begin{gathered} A=2x(140-2x) \\ A=2x(140)-2x(2x) \\ A=280x-4x^2 \end{gathered}`

1.4.2 Given that Area's formula is a quadratic function, its maximum is placed at its vertex. The x-coordinate of the vertex is found as follows:

`x_V=(-b)/(2a)`

where a is the leading coefficient and b is the x-coefficient of the function. In this case, the coefficients are a = -4 and b = 280, then:

`\begin{gathered} x_V=(-280)/(2(-4)) \\ x_V=(-280)/(-8) \\ x_V=35 \end{gathered}`

Evaluating the Area at x = 35, the maximum Area is:

`\begin{gathered} A=280(35)-4(35)^2 \\ A=4900\text{ m}^2 \end{gathered}`