# A rectangle with perimeter of 100 has area at least of 500, within what bounds must the length of the rectangle lie?

Problem The problem states that there is a rectangle that has a perimeter of $100$ and an area of at least $500$ and it asks for the bounds of the length which can be given in interval notation or in the <> (greater than or less than) signs

My steps and thought process So I set some few equations 1)$2x+2y=100$ which becomes $x+y=50$

2)$xy \geq 500$

3)I then made $y = 50-x$ so that I can substitute it into equation #2 to get: $50x-x^2 \geq 500$ which eventually got me $0 \geq x^2-50x+500$ and this is where I got stuck.

• When is a quadratic expression negative ? – Claude Leibovici Jul 7 '16 at 6:21
• The larger root of $x^2-50x+500$ will be useful. – André Nicolas Jul 7 '16 at 6:51
• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Jul 7 '16 at 11:03

You are correct that $x + y = 50$ and that $xy \geq 500$. We can solve the inequality by completing the square. \begin{align*} xy & \geq 500\\ x(50 - x) & \geq 500\\ 50x - x^2 & \geq 500\\ 0 & \geq x^2 - 50x + 500\\ 0 & \geq (x^2 - 50x) + 500\\ 0 & \geq (x^2 - 50x + 625) - 625 + 500\\ 0 & \geq (x - 25)^2 - 125\\ 125 & \geq (x - 25)^2\\ \sqrt{125} & \geq |x - 25|\\ 5\sqrt{5} & \geq |x - 25| \end{align*} Hence, \begin{align*} -5\sqrt{5} & \leq x - 25 \leq 5\sqrt{5}\\ 25 - 5\sqrt{5} & \leq x \leq 25 + 5\sqrt{5} \end{align*}
Alternatively, note that since the perimeter of the rectangle is $100$ units, the average side length is $100/4 = 25$ units. Hence, we can express the lengths of adjacent sides as $25 + k$ and $25 - k$. Hence, the area is $$A(k) = (25 + k)(25 - k) = 625 - k^2$$ The requirement that the area must be at least $500$ square units means \begin{align*} 625 - k^2 & \geq 500\\ 125 & \geq k^2\\ 5\sqrt{5} & \geq |k|\\ 5\sqrt{5} & \geq k \geq -5\sqrt{5} \end{align*} Thus, for the area of the rectangle to be at least $500$ square units, the length of the longer side of the rectangle must be at most $25 + 5\sqrt{5}$ units and the length of the shorter side of the rectangle must be at least $25 - 5\sqrt{5}$ units. Also, note that the maximum area of $625$ square units occurs when $k = 0$, that is, when the rectangle is a square.
Since $x+y= 100$ and $xy \geq 500$, we have $x(50-x) \geq 500$. Thus $x^2 -50x + 500 \leq 0$. Consequently, $x$ must lie between the roots of the quadratic $x^2-50x+500 = 0$. The roots are $25 \pm 5\sqrt{5}$. Thus $25-5\sqrt{5} \leq x \leq 25+5\sqrt{5}$.