Minimum cost of building wall around rectangular region

Question

A triangular plot is such that two of its sides, of lengths $90\text{m}$ and $60\text{m}$ , are perpendicular to each other. There is a housing complex in a rectangular region within the plot. The area of the rectangular region is $4/9^{\text{th}}$ of the area of the triangular plot. Additionally, two sides of the rectangular region lie on the two perpendicular sides of the triangle, and one vertex is on the hypotenuse. The members of the housing complex want to construct a wall along the perimeter of the rectangular region.

If the cost of construction is $\text{Rs. } 5000/\text{m}$ , what is the MINIMUM possible cost of building the wall?

Accepted Answer

B. Let perpendicular sides be along x-axis ($90$) and y-axis ($60$). The hypotenuse equation is $\frac{x}{90} + \frac{y}{60} = 1$. A vertex of the rectangle on hypotenuse has coordinates $(x, y) = (x, 60(1 - x/90))$. Area of rectangle $= xy = 60x\left(1 - \frac{x}{90}ight) = 60x - \frac{2x^2}{3}$. Area of triangle $= \frac{1}{2} 	imes 90 	imes 60 = 2700$. Area of rectangle $= \frac{4}{9} 	imes 2700 = 1200$. So $60x - \frac{2x^2}{3} = 1200 \implies 2x^2 - 180x + 3600 = 0 \implies x^2 - 90x + 1800 = 0$. $(x - 30)(x - 60) = 0 \implies x = 30	ext{ or } 60$. - If $x = 30$, $y = 60(1 - 30/90) = 40$. Perimeter $= 2(30 + 40) = 140	ext{ m}$. - If $x = 60$, $y = 60(1 - 60/90) = 20$. Perimeter $= 2(60 + 20) = 160	ext{ m}$. Minimum perimeter $= 140	ext{ m}$. Minimum cost $= 140 	imes 5000 = 	ext{Rs. } 700,000$.

Answer

Rs. 777,777

Answer

Rs. 700,000

Answer

Rs. 666,667

Answer

Rs. 766,667

Answer

Rs. 433,333

Minimum cost of building wall around rectangular region

Explanation

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