Question

A right triangle has one vertex at the origin and one vertex on the curve y=e^-x/4 for 1 ≤ x ≤ 5. One of the two perpendicular sides is along the x-axis; the other is parallel to the y -axis. Find the maximum and minimum areas for such a triangle.
Round your answers to three decimal places.

Answer 1

The area of a shape is the amount of space it occupies.

The minimum and the maximum areas of the triangle are: 0.389 and 0.736, respectively

The vertex is given as:

$\mathbf{y = e^{-x/4}, \ 1 \le x \le 5}$

From the question, we have:

• A side is on the x-axis (so, the base is x)
• The other side is parallel to the y-axis (so, the height is y)

So, we have:

$\mathbf{Area = \frac 12 \times Base \times Height}$

This gives

$\mathbf{Area = \frac 12 \times x \times e^{-x/4}}$

Differentiate

$\mathbf{A' = \frac 12 \times e^{-x/4} - \frac 18 \times x \times e^{-x/4} }$

Express 1/2 as 4/8

$\mathbf{A' = \frac 48 \times e^{-x/4} - \frac 18 \times x \times e^{-x/4} }$

Factorize

$\mathbf{A' = \frac 48 \times e^{-x/4} (4 - x)}$

$\mathbf{When\ A' = 0;\ 4 - x = 0}$

So:

$\mathbf{x = 4}$ --- the local maxima

So:

When x = 1, we have:

$\mathbf{Area = \frac 12 \times 1 \times e^{-1/4} = 0.389}$

When x = 4, we have:

$\mathbf{Area = \frac 12 \times 4 \times e^{-4/4} = 0.736}$

When x = 5, we have:

$\mathbf{Area = \frac 12 \times 5 \times e^{-4/4} = 0.716}$

Hence, the minimum and the maximum areas of the triangle are: 0.389 and 0.736, respectively

Read more about areas at:

https://brainly.com/question/11906003