Find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side L = 9 cm if one side of the rectangle lies on the base of the triangle. Round your answer to the nearest tenth.
A) 5.5 cm, 4.4 cm
B) 4 cm, 3.91 cm
C) 4.5 cm, 4 cm
D) 4.5 cm, 3.9 cm
E) 7.5 cm, 2.9 cm
F) 9.5 cm, 3.9 cm

Question

Quizplus · Accepted Answer

To maximize the area of the rectangle, we will want the rectangle to be as large as possible, which means it will have a height equal to half the length of the triangle's side (since the rectangle must fit inside the triangle with one side on the base).

Let the base of the rectangle be x, then its height is $\frac{1}{2}(9-x)$. The area of the rectangle is therefore given by:

$A = x \cdot \frac{1}{2}(9-x) = \frac{1}{2}(9x-x^2)$

To maximize this function, we can take its derivative and set it equal to zero:

$A' = 9 - x$

$9 - x = 0 \Rightarrow x = 9$

However, this solution is outside the bounds of the triangle, so the maximum must occur at one of the endpoints, i.e. when $x=0$ or $x=9/2$. Since $x=0$ corresponds to a rectangle with no area, we choose $x=9/2$, which gives a height of $\frac{1}{2}(9-\frac{9}{2}) = \frac{9}{4}$. Thus, the dimensions of the rectangle are:

$x = \frac{9}{2}$ and $h = \frac{9}{4}$

Rounded to the nearest tenth, we get:

$x \approx 4.5$ and $h \approx 3.9$

which matches choice D.

Find the Dimensions of the Rectangle of Largest Area That