For 2 Question Help Meeee :)

[ricopasa]

1.The base of a rectangle is on the x-axis. The upper left vertex is on the line whose equation is y=2x and the
upper right vertex is on the line y+3x=30. For what value of y will the area of the rectangle be a maximum?


2.The lengths of the sides of an equilateral triangle are increasing at a rate of 5 centimeters per hour. At what
rate is the area of the triangle increasing when the side is 10 centimeters long?

Thanks...

 

[pappus]

1.The base of a rectangle is on the x-axis. The upper left vertex is on the line whose equation is y=2x and the
upper right vertex is on the line y+3x=30. For what value of y will the area of the rectangle be a maximum?


2. ...

Thanks...

1. Draw a sketch!

2. The area of the rectangle is \(\displaystyle A = 2x \cdot w\)

3. The upper right vertex of the rectangle has the coordinates \(\displaystyle (x+w, -3(x+w)+30)\)
Therefore
\(\displaystyle 2x = -3(x+w)+30~\implies~\boxed{x=-\frac35w+6}\)

Plug in this term for x into the equation of the area and you'll get a function A wrt w.

4. \(\displaystyle A(w)=\left(-\frac35 w +6 \right) \cdot w\)

is a quadratic function. Calculate the maximum area by determining the coordinates of the vertex.

 

[HallsofIvy]

For (2), let the lengths of the sides be a, a, and b. Dropping a perpendicular from the vertex joining the two sides of length a to the third side gives two right triangles with hypotenuse of length a and one leg of length b/2. Use the Pythagorean theorem to find the length, h, of the other leg of the right triangle, an "altitude" of the original isosceles triangle. That allows you to write the area of the isosceles triangle, (1/2)bh, in terms of a and b. Find the derivative of the area, A, with respect to time, t: \(\displaystyle \frac{dA}{dt}= \frac{\partial A}{\partial b}\frac{db}{dt}+ \frac{\partial A}{\partial a}\frac{da}{dt}\)