Answer to Question #277731 in Calculus for Mamsy

Question

A farmer wants to determine the dimensions of the largest rectangular area tgat can be inscribed in a right angled triagle field with a height h=4 meters and a hypotenuse of 5 meter

Find the dimensions of the rectangle with the maximun area

Answer 1

Given the hypotenuse $c=5$ m, one leg $b=4$ m, Then by the Pythagorean Theorem the second leg is

$$a=\sqrt{c^2-b^2}=\sqrt{5^2-4^2}=3(m)$$

Let $x=$ the width of the rectangle, $y=$ its height.Then

$$\dfrac{4-y}{x}=\dfrac{4}{3}$$

$$4x=12-3y$$

$$x=3-\dfrac{3}{4}y$$

The area of the rectangle will be

$$A=xy$$

Substitute

$$A=A(y)=(3-\dfrac{3}{4}y)y=3y-\dfrac{3}{4}y^2 , 0< y< 4$$

Find the first derivative with respect to $y$

$$A'(y)=3-\dfrac{3}{2}y$$

Find the critical number(s)

$$A'(y)=0=>3-\dfrac{3}{2}y=0=>y=2$$

If $0<y<2, A'(y)>0, A(y)$ increases.

If $2<y<4, A'(y)<0, A(y)$ decreases.

The function $A(y)$ has a local maximum at $y=2.$

Since the function $A(y)$ has the only extremum, then the function $A(y)$ has the absolute maximum for $0<y<4$ at $y=2.$

$$x=3-\dfrac{3}{4}(2)=\dfrac{3}{2}(m)$$

The rectangle $1.5m\times 2m$ will have the maximum area.