The inscribed rectangle of the triangle results in a similar triangle towards the vertex of the of the altitude of the given triangle.
Correct Response;
- The area of the largest rectangle is 24 square feet.
Method used to derive the above response;
Length of the base of the triangle = 12 ft.
Altitude of the triangle = 8 ft.
Required:
The area of the largest rectangle that can be inscribed in the triangle.
Solution:
Let x represent the width of the rectangle, let y represent the height, and let the base of the triangle be on the horizontal x-axis.
From the attached drawing of the situation, we have;
ΔABC is similar to ΔADE
Which gives;
[tex]\displaystyle \frac{8 - y}{8} = \mathbf{\frac{x}{12}}[/tex]
12·(8 - y) = 8·x
[tex]\displaystyle 8 - y = \frac{8 \cdot x}{12} = \frac{2}{3} \cdot x[/tex]
[tex]\displaystyle y = \mathbf{ 8 - \frac{2}{3} \cdot x}[/tex]
Area of the rectangle, A = x × y
Therefore;
[tex]\displaystyle The \ function \ for \ the \ area \ of \ the \ rectangle, \, A = \displaystyle x \times \left( 8 - \frac{2}{3} \cdot x\right) = \mathbf{8 \cdot x - \frac{2}{3} \cdot x^2}[/tex]
The leading coefficient of the quadratic function for the area is negative, therefore, at the maximum area, we have;
- [tex]\displaystyle \frac{d}{dx} (A_{max} ) = 0 = \frac{d}{dx} \left( 8 \cdot x - \frac{2}{3} \cdot x^2 \right) = \mathbf{8 - \frac{4}{3} \cdot x}[/tex]
Which gives;
[tex]\dfrac{4}{3} \cdot x = 8[/tex]
x = 8 × 3 ÷ 4 = 6
The width of the rectangle that gives the maximum area, x = 6 feet
[tex]\displaystyle y = 8 - \frac{2}{3} \cdot x = 8 - \frac{2}{3} \times 6 = 4[/tex]
The height of the rectangle when the area is maximum, y = 4 feet
- The largest area of the rectangle, [tex]A_{max}[/tex] = 6 ft. × 4 ft. = 24 ft.²
Learn more about finding the maximum value of a function here:
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