Optimization problem without calculus

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mickey222

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I find these rather difficult and easy to mess up. I will show you my work, and perhaps someone could go over it and tell me whether it is sound or not. There were a lot of special symbols that I don't know how to enter on this webpage, so I used a math notepad and took screenshots of my work.

The problem is: There is a rectangle with an equilateral triangle on top of it. The width of the rectangle and the base of the triangle are equivalent. The perimeter of the shape is 243 ft. What dimensions will maximize the area of this shape? And what is the max area of the shape?

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mickey222 said:
I find these rather difficult and easy to mess up. I will show you my work, and perhaps someone could go over it and tell me whether it is sound or not. There were a lot of special symbols that I don't know how to enter on this webpage, so I used a math notepad and took screenshots of my work.

The problem is: There is a rectangle with an equilateral triangle on top of it. The width of the rectangle and the base of the triangle are equivalent. The perimeter of the shape is 243 ft. What dimensions will maximize the area of this shape? And what is the max area of the shape?

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Now go back and check whether your initial equations:

2y + 3x = 243 is satisfied.

since the measurement of perimeter had 3 significant figures (243) I would present the results in 3significant figures.
 
Make sure in your Area equation that the second x^2 is NOT under the square root sign. I think you got the wrong max point I found a max at x=56.936. I don't think that the 40.64 is correct. Also since you graphed your area equation then you should already be able to see the max area on your graph without doing any more calculating.
 
You start very well.

[MATH]\text {base of equilateral triangle} = x = \text {one side of rectangle.}[/MATH] Great.

[MATH]\text {height of equilateral triangle} = h.[/MATH] Perfect.

[MATH]\text {other side of rectangle} = y.[/MATH] Just humming along.

[MATH]\text {perimeter} = p = 3x + 2y.[/MATH] Yes Indeed.

[MATH]\text {area} = xy + \dfrac{1}{2} \cdot hx = x \cdot \left ( y + \dfrac{h}{2} \right ).[/MATH] Logical, neat.

[MATH]243 = p = 3x + 2y \implies y = \dfrac{243 - 3x}{2}.[/MATH] First year stuff.

[MATH]h = \sqrt{x^2 - \left ( \dfrac{x}{2} \right )^2} = \sqrt{ \dfrac{3x^2}{4}} = \dfrac{\sqrt{3} \cdot x}{2}.[/MATH] OK

[MATH]\therefore a = x \cdot \left ( \dfrac{243 - 3x}{2} + \dfrac{\sqrt{3} \cdot x}{4} \right ) = \dfrac{486}{4} \cdot x + \dfrac{\sqrt{3} - 6}{4} \cdot x^2.[/MATH]
Your mechanics are fine.

BUT is y a quadratic function of x? Not at all. It is a linear function of x. Are you trying to maximize y? No. What you are trying to maximize is a, and a is a quadratic function of x. This is the danger of formulas; you memorize the formula and fail to think.
 
JeffM said:
BUT is y a quadratic function of x? Not at all. It is a linear function of x. Are you trying to maximize y? No. What you are trying to maximize is a, and a is a quadratic function of x. This is the danger of formulas; you memorize the formula and fail to think.
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Would you mind elaborating on this point a bit further? I understand your remark about formulas and I've experienced it myself many times. I thought that in my work above, however, I did define Y as a linear function of X.
 
AmandasMathHelp said:
Make sure in your Area equation that the second x^2 is NOT under the square root sign. I think you got the wrong max point I found a max at x=56.936. I don't think that the 40.64 is correct. Also since you graphed your area equation then you should already be able to see the max area on your graph without doing any more calculating.
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Thanks for catching that. I also got a max of 56.93600.

To get my other length I plugged that into 2y+3x=243..... so 2(56.93600)+3x=243---> x=(43.04266


this satisfies my equation 2y+3x=243. So now I have my length and width, or x and y.

To find height, plugged 54.93600 into my height equation, and got 49.30802.

To find max area I did: area=(43)(56)+1/2(56)(49) (I entered the long decimals...) and got: 3780


w=56.93600
l=43.04266
h=49.40802

Max area: 3780
 
mickey222 said:
Would you mind elaborating on this point a bit further? I understand your remark about formulas and I've experienced it myself many times. I thought that in my work above, however, I did define Y as a linear function of X.
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I like intelligent questions.

You did not initially define y to be a linear function of x; you demonstrated that y is a linear function of x. Excellent.

However, the formula that you used, namely [MATH]x = - \dfrac{b}{2a}[/MATH],

has nothing to do with linear equations. The formula pertains to quadratic equations. We get that formula after specifying

[MATH]y = ax^2 + bx + c,\ a \ne 0,[/MATH]
and asking at what value of x is the local extremum of y (minimum or maximum as the case may be).

But, given the way you have defined your variables, y is not a quadratic function, and the formula applies to area, which you have defined as A, which is not the same variable as a in the formula you have learned.

So when you say “I plug that value into y” you just jumped off the rails entirely.
 
mickey222 said:
I understand Jeff, thank you for going further into detail. Interestingly, when you plug X into the y equation that I provided, it of course does not satisfy the original perimeter equation, because you get an incorrect value for the reasons you stated. So I plugged X into the original perimeter equation instead to then solve for Y.

It makes far more sense to contextualize the concept by visualizing the shape of this thing also... it basically looks like a two dimensional house. There is no reason that Y would be a quadratic function of X... It would be linear, with the constraints provided.

These things don't often come to me immediately. Did all of this "click" the first time you guys learned about these concepts, or did you have to wrestle with it?
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No. Much of it did not click immediately. That is why exercises are so important. You have to use these ideas repeatedly until you truly understand them. Aristotle called humans reasoning animals, but they don’t reason well, just a lot better than any other species.

There was a lot going on in this problem. There was an opportunity to get mixed up on which variables were which. There is the fact that you are permitted to use graphing, which makes it hard to check your arithmetic. The numbers used in the problem are ugly.
 
JeffM said:
No. Much of it did not click immediately. That is why exercises are so important. You have to use these ideas repeatedly until you truly understand them. Aristotle called humans reasoning animals, but they don’t reason well, just a lot better than any other species.

There was a lot going on in this problem. There was an opportunity to get mixed up on which variables were which. There is the fact that you are permitted to use graphing, which makes it hard to check your arithmetic. The numbers used in the problem are ugly.
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That is good to hear. Yeah, the graphing, I've found, really doesn't do me any favors... And yes, after all of that, I realized I had confused X for Y, which made it a nightmare to back and fix things.
 
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