90% Students Failed To Solve This Geometry Problem

How to find the maximum area of the rectangle inside a triangle
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[@jmlfa]

if PQ=x and QS=y, then y/4=(5-x)/5; xy= (20x - 4x^2)/5 . Looking for d(xy)/dx = 0 leads to x=2.5 and y=2.

[@marioalb9726]

When ratio between sides of rectangle is 5/4, then its area is the maximum: x/y = 5/4 --> y= 4/5 x Similarity of triangles: y/(5-x)=(4-y)/x x.y = (4-y)(5-x) (4/5 x).x = (4 - 4/5 x)(5-x) 4/5 x² = 20 - 8x + 4/5 x² 8 x = 20 x = 2,5 cm ; y = 2 cm A = x.y = 5 cm² ( Solved √ )

[@ناصريناصر-س4ب]

Let's assume that x and y are the lengths of the rectangle. According to Thales' theorem, we have 5x+4y=20. Therefore, the sum of the numbers 5x and 4y is constant, so their product is maximum if they are equal, i.e. 5x=4y=10. Therefore, x=2 and y=5/2, so xy=5.

[@soli9mana-soli4953]

Nice!

[@amritpatel3794]

Use calculus to find Maximum of a function. That will make this puzzel simple. Area of the rectangle = 4X - 4X^2/5. X is horizontal dimention of the rectangle. The maximum of the area will occur when X = 5/2.

[@marioalb9726]

When rectangle es a square x², its area is almost the maximum: Similarity of triangles: x/(5-x)=(4-x)/x x² = (5-x)(4-x) x² = 20-9x+x² 9x=20 x = 20/9 x² = 4,938 cm²

[@marioalb9726]

When ratio between sides of rectangle is 5/4, then its area is the maximum: x/y = 5/4 --> y= 4/5 x Similarity of triangles: y/(5-x)=(4-y)/x x.y = (4-y)(5-x) (4/5 x).x = (4 - 4/5 x)(5-x) 4/5 x² = 20 - 8x + 4/5 x² 8 x = 20 x = 2,5 cm ; y = 2 cm A = x.y = 5 cm² ( Solved √ )

[@imetroangola4943]

Solução: Sejam PB=b e PQ=a. Sendo OQSB um retângulo, temos que QS=b e BS=a. Assim, AP= 4-a e SC=5-b. Como ∆APQ ~ ∆QSC, então: (4-a)/a=b/(5-b) 20-5a-4b+ab=ab →b=(20-5a)/4 [PQSB]=ab=(20a-5a²)/4 A equação f(a)= (20a-5a²)/4 é um gráfico de uma parábola, cujo ponto máximo da parábola se encontra no vértice, logo a=-20/-10=2, isso implica que o valor máximo [PQSB] é *[PQSB]_max = (20×2-5×2²)/4 = 5*

[@jarikosonen4079]

6:28 Also finding max of 5a*4*(1-a) should work.

[@murdock5537]

φ = 30° → sin⁡(3φ) = 1; ∆ ABC → AB = AP + BP = (4 - y) + y; BC = BS + CS = x + (5 - x) AC = AQ + CQ = √41; sin⁡(ABC) = sin⁡(BSQ) = sin⁡(SQP) = 1 → BCA = PQA = δ → tan⁡(δ) = 4/5 = (4 - y)/x → y = 4 - 4x/5 → xy = g(x) = 4x - 4x^2/5 → dg(x)/dx = 4 - 8x/5 = 0 → x = 5/2 → y = 2 → xy_max = 5

[@gelbkehlchen]

Solution: x = PQ = BS, y = BP = SQ. Similarity: x/(5-x) = (4-y)/y |*y ⟹ y*x/(5-x) = 4-y |+y ⟹ y*x/(5-x)+y = 4 ⟹ y*[x/(5-x)+1] = 4 ⟹ y*(x+5-x)/(5-x) = 4 ⟹ y*5/(5-x) = 4 |*(5-x)/5 ⟹ (1) y = 4/5*(5-x) = 4-4/5*x A = Area of ​​the rectangle = x*y [with (1)] = x*(4-4/5*x) = 4x-4/5*x² = -4/5*x²+4x = -4/5*(x²-5*x+2.5²-2.5²) = -4/5*[(x²-5*x+2.5²)-2.5²] = -4/5*[(x-2.5)²-6.25] = -4/5*(x-2.5)²+5 A is a parabola that opens downwards and whose vertex is S = (2.5;5). This is then the maximum value of this area function A. It is at x = 2.5 and the area is then 5 area units.

[@giuseppemalaguti435]

x=base,y=altezza...risulta 4:5=y:(5-x)...y=4-4x/5...A=xy=4x-(4/5)x^2...A'=4-(8/5)x...con A'=0 risulta xmax=20/8,ymax=2...Amax=(20/8)*2=5...ovviamente è un max perché A"=-(8/5)

[@michaeldoerr5810]

Thevmaximum area is 1/4. By golly this is first time in a while that a geometry problem involving maximum area appeared. I am actually wondering there are more geomtry problems that DO involve maximums.

[@imetroangola4943]

Linda solução! 🎉🎉🎉

[@상호정-p9q]

y=-(4/5)x+4, area=x*y=-(4/5)x^2+4x(포물선) max area는 x=2.5일 때, 5

[@Generalist18]

Solved by calculus is that fine?

[@MathBooster]

Yes

[@RealQinnMalloryu4]

(4AB)^2+(5BC)^2={16AB^2+25BC^2}=41ABBC^4 180°PQ/41AVBC^4 =4 .20ABBCPQ^4 4.2^10ABBCPQ^4 4.2^2^5ABBCPQ^4 4.1^1^1ABBCPQ^4 2^2.ABBCPQ^2^2 1^1.ABBCPQ^1^2 ABBCPQ^1^2 (ABBCPQ ➖ 2ABBCPQ+1).

[@imetroangola4943]

I hope one day you explain your crazy math!

[@ยี่สิบเก้าพฤศจิกา]

(4×5)/4 =5 square units