# Math Help - Optimizing Area-HELP!

1. ## Optimizing Area-HELP!

Your parents are going to knock out the bottom of the entire length of the south wall of their house and turn it into a green house by replacing some bottom portion of the wall by a huge sloped piece of glass (which is expensive). They have already decided they are going to spend a certain fixed amount. The triangular ends of the greenhouse will be made of various materials they already have lying around.
The floor space in the greenhouse is only considered usable if they can both stand up in it, so part of it will be unusable, but they don't know how much. Of course this depends on how they configure the greenhouse. They want to choose the dimensions of the greenhouse to get the most usable floor space in it, but they are at a real loss to know what the dimensions should be and how much usable space they will get.

2. This is a bit like all those ladder-to-wall-across-a-fence problems, but you're minimising something other than the ladder.

Start by relabelling the nice picture at http://www.mathhelpforum.com/math-he...dder-wall.html. Let's use capitals for constants.

Z = length of the ladder / glass roof

H = height of the fence / tallest person.

x = distance of roof-end from wall

y = height of other roof-end up wall, but

$y = \sqrt{Z^2 - x^2}$

and finally, u = usable floor.

Use similar triangles to see that the ratio of (x - u) to H is the same as that of x to y.

Express u as a function of x, with constants Z and H. Differentiate u with respect to x and set to zero to find the best x.

Just in case a picture helps differentiate...

... where

... is the chain rule, here wrapped inside...

... the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

In the bottom row, the fourth balloon along (second from left in the big network) is unchanged on its way down from the top row (it wasn't its turn - by the product rule - to get differentiated) but I've tweaked it for the sake of a common denominator.

Then set the derivative to zero. You can thus offer your parents an equation involving x = the optimum extension from the wall, and H and Z for which (presumably) they know the values. I doubt whether we can solve algebraically for x in terms of H and Z... but I'm not at all sure.

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Don't integrate - balloontegrate!

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3. Originally Posted by iheartthemusic29
Your parents are going to knock out the bottom of the entire length of the south wall of their house and turn it into a green house by replacing some bottom portion of the wall by a huge sloped piece of glass (which is expensive). They have already decided they are going to spend a certain fixed amount. The triangular ends of the greenhouse will be made of various materials they already have lying around.
The floor space in the greenhouse is only considered usable if they can both stand up in it, so part of it will be unusable, but they don't know how much. Of course this depends on how they configure the greenhouse. They want to choose the dimensions of the greenhouse to get the most usable floor space in it, but they are at a real loss to know what the dimensions should be and how much usable space they will get.

1. Let G denote the length of the glas pane (because the price for the glas is fixed G is a constant), P the length of the longest person, b the part of the wall which has to be taken off, a the complete floor and x the usable part of the floor. (see attachment)

2. Define a coordinate system with the floor on the x-axis and the wall on the y-axis. G is a segment of the line

$y=-\frac ba x+b$

3. According to Pythagorean theorem you get:

$b^2+a^2 = G^2~\implies~a = \pm \sqrt{G^2-b^2}$
That means the equation of the line becomes:

$y=-\frac b{\sqrt{G^2-b^2}} x+b$ (Since a is pointing to the right I took only the positive value of a)

4. Now y = P (which is for a certain period of time a constant)

$P=-\frac b{\sqrt{G^2-b^2}} x+b~\implies~x=\dfrac{(b-P) \sqrt{G^2-b^2}}{b}$

This is an equation of the function x(b). Differentiate x wrt b and solve for b the equation x'(b) = 0.

5. I've got:

$x'(b)= \dfrac{G^2 \cdot P-b^3}{b^2 \sqrt{G^2-b^2}}$

and

$b = \sqrt[3]{G^2 \cdot P}$

6. Plug in this value into x(b) - and you'll get a really nasty looking term for x.