# Maximum area of a rectangle whose perimeter is $100$. (Derivative Approach)

Question:

Find the maximum area of a rectangle whose perimeter is $$100$$?

My Approach:

(I am using derivative method)

Some Basic rules:

1.) If $$f''(x) > 0$$ then function has minimum value at $$x$$

2.) If $$f''(x) < 0$$ then function has maximum value at $$x$$

Let the length of rectangle $$= x$$

and the width of rectangle $$= y$$

$$2(x+y)=100$$

$$y = 50 - x$$

As Area $$= x × y$$

Put the value of y in above equation

$$f(x) = Area = x(50 - x)$$

$$f(x) = 50x - x^2$$

Taking first derivative: $$f'(x) = 50 - 2x$$

Assuming the first derivative to be equal to zero we get the value of x which is $$x = 25$$.

Now taking double derivative of above equation and putting the value of $$x$$ we get $$f''(x) = -2$$ which is less then zero so the function of Area is maximum hence the Area of rectangle will also be maximum.

Now what approach should i use to calculate Area?

Conclusion:

Is there a more neat way doing the above question more quickly?

What if the question says to calculate minimum Area?

Help would be appreciated.

Thanks,

(Sorry for bad english and if someone finds any mistake feel free to edit and correct them :) )

We have

$$P = 2(l+w) \implies 100 = 2(l+w) \iff \color{blue}{50 = l+w} \tag{1}$$

$$\color{green}{A = lw} \tag{2}$$

Rewriting $$(1)$$ in terms of $$w$$ (you could do so in terms of $$l$$), you’d get

$$w = 50-l$$

Plugging $$w = 50-l$$ in $$(2)$$, you have

$$A = l(50-l)$$

$$\color{purple}{A = 50l-l^2} \tag{3}$$

You’ve gotten it correctly until here. Treat $$A$$ as a function. The resulting quadratic is concave down, so the vertex is a maximum.

$$l_{vertex} = \frac{-b}{2a} = \frac{-50}{2(-1)} = 25$$

Plugging $$l = 25$$ in $$(1)$$, it becomes clear that $$w = 25$$. Hence, the maximum area occurs when there is a square, so

$$\boxed{A = s^2 = 25^2 = 625}$$

Using only quadratics is faster than using optimization, but that is of course correct as well:

$$\frac{dA}{dl} = 50-2l$$

$$\frac{dA}{dl} = 0 \implies 50-2l = 0 \implies l = 25$$

from which the same answer is obtained.

In response to your second question, you can’t find a minimum area, because it simply doesn’t exist. (Unless if you set $$w = 0$$ in which case the area becomes $$0$$.) A concave down parabola has a maximum, not a minimum. A minimum point is found when the parabola is concave up, which isn’t the case here.

• What is Quadratics Method? Its the first time i heard of it, could you please explain? – Gingitsune Dec 30 '18 at 12:44
• Do you know about quadratic equations? (You probably do, I suppose.) If yes, you can treat the derived equation for area as a quadratic function. Here, $A = 50l-l^2$, so the coefficient of $l^2$ is negative. This means the vertex of the graph is a maximum. All you have to do is find where it occurs, and you can use the formula $h = \frac{-b}{2a}$ to do that. You can also use optimization and calculus, but it’s not necessary. – KM101 Dec 30 '18 at 12:47
• What if the coefficent of $l^2$ is positive? – Gingitsune Dec 30 '18 at 12:58
• Since area is $\ge 0$ by definition, the minimum area is 0, obtained for the degenerate rectangle when one of its sides is 0 and the other is 50. Minima (and maxima) can occur not only in the interior of a region (where setting a derivative equal to 0 can find them) but also on the boundaries. – NickD Dec 30 '18 at 12:59
• @Malik Talha That would be the case for an optimization problem in which you’re looking for a minimum. You can still use the same approach. Either find where the vertex of the parabola occurs or differentiate and set the derivative to $0$ (and solve from there). – KM101 Dec 30 '18 at 13:02

Let $$a$$ and $$b$$ be sides-lengths of adjacent sides of the rectangle.

Thus, by AM-GM $$ab\leq\left(\frac{a+b}{2}\right)^2=\left(\frac{50}{2}\right)^2=625.$$ The equality occurs for $$a=b$$, which says that we got a maximal value.

• Sorry i couldn't understand your approach would you elaborate please? – Gingitsune Dec 30 '18 at 12:42
• @Malik Talha What exactly? $ab$ it's an area of the rectangle and $2(a+b)=100.$ – Michael Rozenberg Dec 30 '18 at 13:43

The area is, as you say, $$A(x)=x(50-x)$$, a quadratic with roots $$x=0$$ and $$x=50$$. The graphic is a parabola simmetric respect to the line $$x=\frac{0+50}{2}=25$$. Also, the parabola is concave down because the principal coefficent is negative. Then, the maximum is when $$x=25$$ and the value is $$25(50-25)=25^2$$.

You don't need calculus here, though it does work.

You can do this neatly by noting that $$(l+w)^2-(l-w)^2=4lw$$

You want to maximise $$lw$$ and $$l+w$$ is fixed at $$50$$ so you need $$l-w$$ as small as possible.

To minimise the area you clearly want $$l+w$$ and $$l-w$$ to be as close to one another as possible.