0
$\begingroup$

I'm working through some optimization questions and I got stuck, and ended up with the wrong answer. The question is:

There's a garden that needs to be enclosed, you have $60$m of fencing, one side is against a house so you need to enclose only three sides, maximize the area.

So what I did was,

Constraint: $2x+y=60$

Maximize area = $xy$

$$y = 60 - 2x$$ $$f(x) = x(60-2x)$$ $$= 60x-2x^2$$

$$f'(x) = 60-4x$$ $$= 4(15-x)$$

Now this is the part where I'm confused, I know that $y = 60-2x$, but how do I solve for $x$? I used $x = 15$, which gave me the wrong answer.

$\endgroup$
  • $\begingroup$ Why do you think that x=15 is the wrong answer? $\endgroup$ – DJohnM Apr 4 '13 at 3:21
  • $\begingroup$ The shape is restricted to rectangles, I assume. Another way to verify is to consider that $(2x)y$ is being maximised, while $(2x)+y$ is fixed. This means $(2x)=y$ at maximum. Again gives $x=15$. $\endgroup$ – Macavity Apr 4 '13 at 3:37
1
$\begingroup$

Your work is correct but not complete.

The function for area starts at $0$, for $x=0$, and is positive for $x>0$. Until $x>30$, where the area becomes negative.

Setting the derivative, $60-4x=0$, gives one point, $x=15$, where the area must be a maximum. Substituting, the area is $15\times30=450m^2$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.