Maximum value of area of rectangle

Question

Tried attempting by using altitude and similarity of triangles, but the problem is that variables are not getting elliminated.

Answer 1

By Cavalieri's principle we can make the outer triangle isosceles without changing the rectangles' area. By hyperbolic transformations we can make this triangle have base 6 and height 4, again keeping the rectangle areas intact.

Let $p=P_1S_1$ and $q=P_2S_2$. The triangle area outside the rectangles, which we want to minimise, is $$p\cdot\frac34p+q\cdot\frac34q+(4-p-q)\cdot\frac34(4-p-q)$$ (The $\frac34$ factors are there because we can rearrange the unoccupied space into three 4:3 rectangles whose longer sides are $p,q,4-p-q$.) $$=\frac34\left(p^2+q^2+(4-(p+q))^2\right)$$ $$=\frac32(p^2+pq+q^2-4p-4q)+12$$ The gradient of the expression in brackets is $(2p+q-4,2q+p-4)$ and this is zero when $p=q=\frac43$. Thus these are the rectangle heights maximising the area occupied, which is $$\frac32\left(8\cdot\frac43-3\cdot\frac43\cdot\frac43\right)=8$$

Answer 2

Let's call $x$ the height from $A$ onto $S_2R_2$, $y=|S_2P_2|$, and $z=|S_1P_1|$. The area of rectangles is then $$A_r=z|S_1R_1|+y|S_2R_2|$$ We can write these segments in terms of $|BC|$ by looking at similar triangles: $$|S_2R_2|=|BC|\frac{x}{x+y+z}\\|S_1R_1|=|BC|\frac{x+y}{x+y+z}$$ Therefore: $$A_r=|BC|\frac{xy+yz+zx}{x+y+z}$$ Since $|BC|$ and $x+y+z$ are constant for a given triangle, we want to maximize $xy+yz+zx$ with the constraint that $x+y+z=k$, where $k$ is some constant. Using Lagrange multiplier method: $$\partial_x (xy+yz+zx-\lambda(x+y+z-k))=0\\\partial_y (xy+yz+zx-\lambda(x+y+z-k))=0\\\partial_z (xy+yz+zx-\lambda(x+y+z-k))=0$$ you get: $$y+z-\lambda=0\\x+z-\lambda=0\\x+y-\lambda=0$$ Using $x+y+z=k$, when adding these equations you get$$2k-3\lambda=0$$ or $\lambda=\frac{2}{3}k$. The solution is $$x=y=z=\frac{k}{3}$$ Then $$\max(A_r)=\frac{3(k/3)^2}{k}|BC|=\frac{k|BC|}{3}$$ Notice that $k|BC|$ is twice the area of the triangle, so $$\max(A_r)=8$$

Answer 3

The two apiled rectangles leave an area associated to three similar triangles. The rectangles will have maximum area when the three similar triangles are equal

or ressuming

$$ \frac 12 b\cdot h = 12 \Rightarrow b\cdot h = 24\\ 3\frac 12 \left(\frac b3\cdot\frac h3\right) = 4 $$

so the maximum area is $12-4 = 8$

NOTE

The problem is the same depending on the number of apiled squares. In the case of $n$ we have the maximum squares area is given by $12 - n\left(\frac{1}{2}\frac{b}{n}\frac{h}{n}\right) = 12-\frac{12}{n}$ for $n = 1, 2,\cdots$

To show the necessity of the equality between the small excess triangles think on the problem depicted in the included picture.

Answer 4

Drop a perpendicular from $A$ to the base, and consider the "right half" $S$ of the triangle, whereby we may assume $A=(0,1)$, $C=(1,0)$, $Q_1=(x,0)$, $\>0\leq x\leq1$. If there were just one rectangle we would have to maximize $x(1-x)$, leading to $x={1\over2}$. For two rectangles we therefore have to maximize $$f(x):=x(1-x)+\left({x\over2}\right)^2=x-{3\over4}x^2\ .$$ The maximum is at $x={2\over3}$ with $f\left({2\over3}\right)={1\over3}$. It follows that $${{\rm area}_\max(R_1\cup R_2)\over{\rm area}(S)}={f\bigl({2\over3}\bigr)\over{1\over2}}={{1\over3}\over{1\over2}}={2\over3}\ ,$$ so that the answer to the question is ${2\over3}\cdot12=8$.