A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single strand of electric fence. With 1400m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions?
Is there a way to solve this without Calculus? This sounds like a simple perimeter problem. If you have to use Calculus to solve this, how would I start? Could someone show me the steps?
$\endgroup$2 Answers
$\begingroup$Let $r$ be the length of the rectangle along the river and $s$ be the length of the other side. Thus, we have to:
Max $rs$
Subject to:
$r + 2s = 1400$
Substitute for $r$ back into the objective function and we get the problem reformulated as:
Max $(1400-2s) s$
You can find the value of $s$ either via calculus or by completing the square.
$\endgroup$ 1 $\begingroup$Yes, you can solve this without calculus by a process called quadratic optimisation (though the math behind it really uses calculus, you can do it without taking a single derivative or integral).
The plot is rectangular, so we have four sides which sum to 1400: $x + 2z = 1400$
Area $A = x \cdot z$
We'll want to state $A$ in terms of x only, so we use a simultaneous equation. Rearrange $x + 2z = 1400$ to form $2z = 1400 - x$, simplify to $z = 700 - \frac{x}{2}$
Plug that into the area equation:
$A = x \cdot (700 - \frac{x}{2})$
$A = 700x - \frac{x^{2}}{2}$
Now we have a quadratic, so we can optimise it. (If we wanted to use calculus, we could see where the derivative is equal to 0, but we won't do that.) We will do this by finding the vertex of the quadratic.
We'll want the $a,b,c$ values of our quadratic in form ($ax^{2} + bx + c$):
$a = \frac{-1}{2}$
$b = 700$
$c = 0$
To find the x-value of the vertex, we use the formula $x=\frac{-b}{2a}$:
$x=\frac{-700}{2\frac{-1}{2}}$
$x=\frac{-700}{\frac{-2}{2}}$
$x=\frac{-700}{-1}$
$x=700$
We just plug that in to find the A-value:
$A = 700x - \frac{x^{2}}{2}$
$x=700$
$A = 700(700) - \frac{700^{2}}{2}$
$A = 245000$
This only leaves the question of what the length of the other 2 sides is. If $x + 2z = 1400$, and $x = 700$, then $z = 350.$
So there's the answer: The area is maximised (to 245000) when one side is 700 units long, and the other two sides are 350 each.
$\endgroup$ 2
