Question

These 2 questions I need background/reasoning on every step of answer as I’ve never done this problem before.

Answer 1

Given:

The length of the rectangle ABCD (in metres)=

$l=2x$

Breadth of the rectangle ABCD (in metres)=

$b=y$

Perimeter of the rectangle ABCD (in metres)=

$280\text{ }$

Required:

(1) To show that the area A, of the rectangle can be written as:

$A=280x-4x^2$

(2) Maximum possible area that the rectangle can occupy.

Answer:

(1) We know that the perimeter(P) of the rectangle is,

$P=2(l+b)$

Therefore, substituting values, we get,

$\begin{gathered} P=2(l+b) \\ 280=2(2x+y) \\ \frac{280}{2}=2x+y \\ 140=2x+y \\ y=140-2x \end{gathered}$

Now, we know that the area(A) of the rectangle is,

$A=l\times b$

Therefore, substituting values, we get,

$\begin{gathered} A=l\times b \\ A=2x\times y \\ A=2x\times(140-2x) \\ A=(2x\times140)-(2x\times2x) \\ A=280x-4x^2 \end{gathered}$

Hence, the area A, of the rectangle can be written as:

$A=280x-4x^2$

(2) The maximum possible area that the rectangle can occupy is,

$\begin{gathered} A=l\times b \\ A=2x\times y \\ A=2x\times(140-2x) \\ A=(2x\times140)-(2x\times2x) \\ A=280x-4x^2 \end{gathered}$

Hence, the maximum possible area of the rectangle is,

$A=280x-4x^2$

Final Answer:

The area A, of the rectangle can be written as:

$A=280x-4x^2$

The maximum possible area of the rectangle is,

$A=280x-4x^2$