Maximization problems I am in a introductory course of calculus in several variables and i have these problems.


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*A farmer wants to build a corral with pentagonal form(not regular) that is formed by the union of a rectangle and a isoscels triangle, as in figure. The farmer has $140$ meters of wire, so the total perimeter must be $140$ meters. Calculate the maximum area(in square meters) for the corral.


Use $x$ and $y$ variables like in figure to find the height of rectangle and the height of triangle.
Define a function $A(x,y):\mathbb{R}^2-> \mathbb{R}$ as the area of the corral.


*A factory use cotton and wool as raw material to produce cloth. The amount of produced cloth is given by    $Q(x,y) = xy + 1$, where $x$ is the amount of wool, and y is the amount of cotton (in kg). One kg of wool costs $7.3$ and one kg of cotton costs $8.4$ . The factory can just spend $10 000$. How many kg of cotton must buy for each kg of wool to produce the maximum amount of cloth? 


Well, for the first problem i put w as the triangle height, so  $(x/2)^2 + w^2 = y^2$ (pythagorean theorem)
then $w = \sqrt{y^2 - (x/2)^2}$. I dont know how to continue, i not even know how to calculate the rectangle height.
(Observation: In the figure there is marked the line separating the triangle from the rectangle but we ignore it in the problem. The corral is only the edge of the pentagonal figure.)
For the second problem from hypothesis i have $7.3x + 8.4y <= 10 000$
In the other hand due to this part "How many kg of cotton must buy for each kg of wool to produce the maximum amount of cloth?" i think we must put y as a function of $x$, i.e. we must find $f$ with $f (x) = y$ and such that $Q (x, f (x))$ be maximum.
But i think maybe the words "How many kg of cotton must buy for each kg of wool"  could mean a ratio like $r = (y / x)$.
 Its all i have. 

How i continue?
Please help me, i am confused. Any hint is welcome.
 A: Let's start from the first: introduce the variable $x,y,z$ in the drawing.
At this point, knowing that the perimeter must be $140$ we get the equation $2x+y+2z=140$ thus we can say that $z=70-x/2-y$.
Now we construct the function of the area:

*

*the rectangle's area is clearly $x\times y$.

*for the triangle we calculate the height: $h=\sqrt{z^2-(x/2)^2}=\sqrt{(70-x/2-y)^2-(x/2)^2}$ from which we deduce the area $\frac12 x\sqrt{(70-x/2-y)^2-(x/2)^2}$
Adding the two together leads:
$$A(x,y)=xy+\frac12 x\sqrt{(70-x/2-y)^2-(x/2)^2}$$
which is a function that you shall minimize, under the condition that $x,y\geq 0$ and $x+2y\leq 140$

For the second, indeed it looks like they are asking for a ratio. But if you think shortly about it, it's clear that to maximise you income you need to use all the money, so you can write $7.3x+8.4y=10000$ (instead of $\leq$).
So the profit function simply becomes univariate:
$$Q(x)=x\times \left(\frac{10000-7.3x}{8.4}\right)+1,$$
which is nothing but a parabola, whose maximum is at the vertex. Once you find $x$ you can calculate $y$ and calculate the ratio $x/y$.
