Maximum area from fixed length fence

[Karol]

Homework Statement

Homework Equations

First derivative=maxima/minima/vertical tangent/rising/falling

The Attempt at a Solution

a are the sides of the base and b is the height
$$A=4ab+a^2,~~V=a^2b=a^2\frac{A-a^2}{4a}=...=\frac{1}{4}a(A-a^2)$$
$$V'=\frac{1}{4}(A-a^2-2a^2)=\frac{1}{4}(A-3a^2)$$
$$V'=0~\rightarrow~a=\sqrt{\frac{A}{3}},~b=\frac{\sqrt{3A}}{4}$$
According to the book b should be $~\displaystyle \frac{1}{2}\sqrt{A/3}$

 

[LCKurtz]

Homework Statement

View attachment 214615

Homework Equations

First derivative=maxima/minima/vertical tangent/rising/falling

The Attempt at a Solution

a are the sides of the base and b is the height
$$A=4ab+a^2,~~V=a^2b=a^2\frac{A-a^2}{4a}=...=\frac{1}{4}a(A-a^2)$$
$$V'=\frac{1}{4}(A-a^2-2a^2)=\frac{1}{4}(A-3a^2)$$
$$V'=0~\rightarrow~a=\sqrt{\frac{A}{3}},~b=\frac{\sqrt{3A}}{4}$$
According to the book b should be $~\displaystyle \frac{1}{2}\sqrt{A/3}$

Your value for $a$ is correct. You need to show your work how you got $b$ for us to find your error.

Also, what does this problem have to do with the title of the thread?

 

[Ray Vickson]

Homework Statement

View attachment 214615

Homework Equations

First derivative=maxima/minima/vertical tangent/rising/falling

The Attempt at a Solution

a are the sides of the base and b is the height
$$A=4ab+a^2,~~V=a^2b=a^2\frac{A-a^2}{4a}=...=\frac{1}{4}a(A-a^2)$$
$$V'=\frac{1}{4}(A-a^2-2a^2)=\frac{1}{4}(A-3a^2)$$
$$V'=0~\rightarrow~a=\sqrt{\frac{A}{3}},~b=\frac{\sqrt{3A}}{4}$$
According to the book b should be $~\displaystyle \frac{1}{2}\sqrt{A/3}$

It looks like your $A$-equation is incorrect: you make the bin by cutting out four corner squares of sides $b$, so those (wasted) squares are also part of the total material area $A$. (The fact that the problem talks about wasted material suggests that you start with a square of area $A$ and then discard parts of it. Your modeled scenario would have no waste, because you would fabricate the material already in the shape of a square with missing corners.)

Anyway, I get a different solution from yours, because my area-equation is different from yours.

Why do you call it a fence?

 

[StoneTemplePython]

For what its worth: a much simpler approach is to wield $GM \leq AM$.

i.e. if you want to maximize:

$x_1 x_2 ... x_n$

subject to $\gamma_1 x_1 + \gamma_2 x_2 + ... + \gamma_n x_n = c$.

where each $x_i \gt 0$, $\gamma_j \gt 0$ and $c\gt 0$ (here the constant $c$ is the 'given amount of material'), application of $GM \leq AM$ to your problem would tell you that, $n= 3$, and

$x_1 = x_2$
(i.e. $x_1$ and $x_2$ are your $a$)

and

$b = x_3 = \frac{1}{2} x_1$.

In general knowing a useful inequality can save you a lot of work and relations between arithmetic means and geometric means come up a lot. And wielding inequalities comes up quite a bit in calc / analysis.

 

[Karol]

The name is, by mistake, from an other problem, sorry.

$$A=a^2+4(b^2+ab)=(a+b)^2+2ab,~~V=a^2b$$
I can't extract a² or b alone from A

 

[Ray Vickson]

The name is, by mistake, from an other problem, sorry.
View attachment 214623
$$A=a^2+4(b^2+ab)=(a+b)^2+2ab,~~V=a^2b$$
I can't extract a² or b alone from A

The problem is quite easy if you use the Lagrange multiplier method. That saves you from having to solve a quadratic equation to get $a$ in terms of $A$ and $b$, or $b$ in terms of $A$ and $a$.

 

[LCKurtz]

I don't see why you would interpret that the box was cut from a square. The OP's solution is correct right up to his last step. He made an error getting $b$ after he had $a$. All that is needed is to correct that algebra, which he hasn't shown us. The book's answer is correct.

 

[ehild]

The problem says "neglect waste in construction". @LCKurtz is right. The OP's solution is correct except the last step.

 

[Karol]

$$A=4ab+a^2~\rightarrow~b=\frac{A-a^2}{4a}$$
$$a=\sqrt{\frac{A}{3}}~\rightarrow~b=\frac{A-\frac{A}{3}}{4\sqrt{\frac{A}{3}}}$$
$$=\frac{2A}{12\sqrt{\frac{A}{3}}}=\frac{A}{6\sqrt{\frac{A}{3}}}=\frac{\sqrt{3A}}{6}$$

 

[Ray Vickson]

The problem says "neglect waste in construction". @LCKurtz is right. The OP's solution is correct except the last step.

Right: I know that is what it says. My interpretation is just different from yours or Kurtz's or the OP's.

In past teaching I have assigned both versions of this problem, but in the scenario modeled by the OP I would just say "for a given area $A$ of sides and bottom, maximize the volume", with no mention at all of "waste".

 

[LCKurtz]

$$A=4ab+a^2~\rightarrow~b=\frac{A-a^2}{4a}$$
$$a=\sqrt{\frac{A}{3}}~\rightarrow~b=\frac{A-\frac{A}{3}}{4\sqrt{\frac{A}{3}}}$$
$$=\frac{2A}{12\sqrt{\frac{A}{3}}}=\frac{A}{6\sqrt{\frac{A}{3}}}=\frac{\sqrt{3A}}{6}$$

So, are you happy with that answer? Do you know whether or not it is correct?

 

[Karol]

$$\frac{2A}{12\sqrt{\frac{A}{3}}}=\frac{A}{6\sqrt{\frac{A}{3}}}=\frac{\sqrt{3A}}{6}=\frac{\sqrt{3A}\sqrt{3}}{6\sqrt{3}}=\frac{1}{2}\sqrt{\frac{A}{3}}$$

 

[LCKurtz]

I will take that as a "yes".