# Thread: Applications of Derivatives - optimization problems

1. ## Applications of Derivatives - optimization problems

This topic of optimization is tough for me to grasp and having lots of trouble with the problems

Question
a rectangle has its base on the x-axis and its upper two vertices on the parabola y = 12-x^2. what is the largest area the rectangle can have, and what are its dimensions ?

I drew a diagram of a parabola going to the negative and then the rectangle in it
but now would I set the area of a rectangle A = xy equal to the equation above?

2. Originally Posted by jddery
This topic of optimization is tough for me to grasp and having lots of trouble with the problems

Question
a rectangle has its base on the x-axis and its upper two vertices on the parabola y = 12-x^2. what is the largest area the rectangle can have, and what are its dimensions ?

I drew a diagram of a parabola going to the negative and then the rectangle in it
but now would I set the area of a rectangle A = xy equal to the equation above?
Ok, so when you draw the diagram you need to try to label the sides of the rectangle. I labeled the vertices on the x-axis as $\displaystyle -x$ and $\displaystyle x$. So the length of the base is $\displaystyle 2x$. The rectangles vertices on the parabola are labeled $\displaystyle y$, so the lenght of the sides is $\displaystyle y$. So the area, is $\displaystyle A(x,y)=2xy$. Now substitue the parabola to make this a function of $\displaystyle x$ alone.

3. Originally Posted by jddery
This topic of optimization is tough for me to grasp and having lots of trouble with the problems

Question
a rectangle has its base on the x-axis and its upper two vertices on the parabola y = 12-x^2. what is the largest area the rectangle can have, and what are its dimensions ?

I drew a diagram of a parabola going to the negative and then the rectangle in it
but now would I set the area of a rectangle A = xy equal to the equation above?
Because of symmetry, the two base vertices are at $\displaystyle (\pm x,0)$ and the two upper vertices are $\displaystyle 12-x^2$. So the area is $\displaystyle \text{base}\times\text{height}=(x-(-x))\cdot(12-x^2)=24x-2x^3$

Take the derivative of this and set it equal to zero to find $\displaystyle x$ that maximizes the area.