Answer to Question #249926 in Calculus for asd

Question

General Application of Derivatives. Draw the necessary figure and indicate the dimension given.

1. A triangular corner lot has perpendicular sides of lengths 120 m and 160 m. Find the area of the largest rectangular building that can be constructed on a lot with sides parallel to the streets.

Answer 1

Let x be the length of a rectangle parallel to the 160 m side

    y be the height of a rectangle parallel to the 120 m side.

As we noticed, If the rectangular building was placed in a triangular lot, we can have 3 similar triangles. The given triangle and the 2 triangles on both sides. A similar triangle has the following ratio:

$\begin{aligned} &\frac{y}{160 m-x}=\frac{120 m}{160 m}=\frac{3 m}{4 m} \\ \Rightarrow &\frac{y}{160 m-x}=\frac{3 m}{4 m} \\ \Rightarrow &4 y=3(160 m-x) \\ \Rightarrow&4 y=480 m-3 x \\ \Rightarrow&y=120 m-\frac{3 m x}{4 m} \end{aligned}$

Since A = xy 

$Then, \quad A=x\left(120-\frac{3 x}{4}\right) \\ =120 x-\frac{3 x^{2}}{4}\\ Get \ the \ maximum \ value \ of \ A \\ \frac{dA}{ d x}=120-\frac{3 x}{2}=0\\ \begin{aligned} &120-\frac{3 x}{2}=0 \\ \Rightarrow&\frac{3 x}{2}=120 \\ &x=80 \end{aligned}\\ From \ y=120-\frac{3 x}{4} , substitute \ the \ value \ of \ \mathrm{x} .\\ \begin{aligned} &y=120-\frac{3(80)}{4} \\ &y=120-60 \\ &y=60 \end{aligned}\\ Since \ x=80 \ and \ y=60$

Therefore the dimension of the largest rectangular building that can be constructed on the triangular lot with the side parallel to the street is 80m by 60m.