# Thread: Word Problem with a Graph

1. ## Word Problem with a Graph

1) What is the area of the largest rectangle below the graph of the function f(x)=4-3x^2 and above the x-axis with its base on the x-axis?
-I am unable to post the graph, but I am able to find the equation of the area to be 2z(4-3z^2). However, I do not know how to continue the problem.

Any help will be much appreciated!

-M

2. $\displaystyle y=4-3x^{2}$

The area of the rectangle is A=2xy

Therefore, $\displaystyle A=2x(4-3x^{2})=8x-6x^{3}$

$\displaystyle A'(x)=8-18x^{2}$

$\displaystyle 8-18x^{2}=0$

$\displaystyle x=\pm{\frac{2}{3}}$

$\displaystyle y=\pm{\frac{8}{3}}$

3. Originally Posted by blurain
1) What is the area of the largest rectangle below the graph of the function f(x)=4-3x^2 and above the x-axis with its base on the x-axis?
-I am unable to post the graph, but I am able to find the equation of the area to be 2z(4-3z^2). However, I do not know how to continue the problem.

Any help will be much appreciated!

-M
f(x) = 4 -3x^2 is "vertical" parabola that opens dowwnward, whose vertex is (0,4), whose axis of symmetry is x=0 or the f(x)=y axis.

So, area A = 2x(y)
A = 2x(4 -3x^2) -------------not z's.

The largest A is when dA/dx = 0, so,
dA/dx = 2x[-6x] +(4 -3x^2)[2]
0 = -12x^2 +8 -6x^2
0 = -18x^2 +8
x^2 = 8/18 = 4/9
x = 2/3

So, max A = 2(2/3)[4 - 3(2/3)^2]
max A = (4/3)[4 - 4/3]
max A = (4/3)[(3*4 -4)/3]
max A = (4/3)[8/3]
max A = 32/9 sq.units ------------answer.
In decimals,
max A = 3.555....sq.units