Math Help - Max/Min Application

1. Max/Min Application

The parabola y = x^2 and the line y = mx +b intersect at the points A (α, α^2) and B (β, β^2) as shown.

a) Explain why αβ = - b

b) Show that AB = √ (m^2 + 4b) (m^2 + 1)

c) The point P(x, x^2) lies on the parabola between A and B. Show that the area of triangle APB is given by:

Area of Triangle APB = ½ (b + mx – x^2) √ (m^2 + 4b)

d) Find the position of point P that maximizes the area of triangle APB in terms of m.

2. Originally Posted by xwrathbringerx

The parabola y = x^2 and the line y = mx +b intersect at the points A (α, α^2) and B (β, β^2) as shown.

a) Explain why αβ = - b

b) Show that AB = √ (m^2 + 4b) (m^2 + 1)

c) The point P(x, x^2) lies on the parabola between A and B. Show that the area of triangle APB is given by:

Area of Triangle APB = ½ (b + mx – x^2) √ (m^2 + 4b)

d) Find the position of point P that maximizes the area of triangle APB in terms of m.

to #a):
Use the 2-point-formula of a straight line:

$\dfrac{y-\alpha^2}{x-\alpha}=\dfrac{\beta^2-\alpha^2}{\beta-\alpha}$

$y-\alpha^2=\dfrac{(\beta^2-\alpha^2)(x-\alpha)}{\beta-\alpha}$

$y=\dfrac{((\beta-\alpha)(\beta+\alpha))(x-\alpha)}{\beta-\alpha}+\alpha^2$

$y=(\beta+\alpha)x-\alpha \beta-\alpha^2+\alpha^2$

$y=(\beta+\alpha)x-\alpha \beta$ Compare this equation with

$y=m \cdot x+b$ which gives $b = - \alpha \beta$

3. Originally Posted by xwrathbringerx

The parabola y = x^2 and the line y = mx +b intersect at the points A (α, α^2) and B (β, β^2) as shown.
...
d) Find the position of point P that maximizes the area of triangle APB in terms of m.

The area of the triangle is a function with respect to x. Calculate the first derivation and set this term equal to zerO.

$A(x)=\dfrac12(b+mx-x^2)\underbrace{\sqrt{m^2+4b}}_{constant!}$

$A'(x)=\dfrac12 \sqrt{m^2+4b} \ \cdot \ (m-2x)$

$A'(x) = 0~\implies~m-2x=0~\implies~x=\dfrac m2$

4. Sorry but how do I find the area of the triangle APB?

5. Originally Posted by xwrathbringerx

The parabola y = x^2 and the line y = mx +b intersect at the points A (α, α^2) and B (β, β^2) as shown.

a) Explain why αβ = - b

b) Show that AB = √ (m^2 + 4b) (m^2 + 1)
b) From part a) you got, $\alpha\beta=-b$

Also, slope of given line (y=mx+b) = slope of AB

$m = \frac{\beta^2-\alpha^2}{\beta-\alpha}=\beta+\alpha$

Now, use distance formula to calculate distance $AB= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AB=\sqrt{(\beta-\alpha)^2+(\beta^2-\alpha^2)^2}$

$AB=\sqrt{(\beta-\alpha)^2+(\beta-\alpha)^2(\beta+\alpha)^2}$

$AB=\sqrt{(\beta-\alpha)^2+(\beta-\alpha)^2m^2}$

$AB=\sqrt{(\beta-\alpha)^2(1+m^2)}$

Now,

$(\beta-\alpha)^2 = \beta^2-2\alpha\beta+\beta^2$

$= \beta^2+2\alpha\beta+\beta^2-4\alpha\beta$

$=(\beta+\alpha)^2-4\alpha\beta$

$=m^2-4(-b)=m^2+4b$

$(\beta-\alpha)^2=m^2+4b$

Therefore,

$AB=\sqrt{(\beta-\alpha)^2(1+m^2)} = \sqrt{(m^2+4b)(1+m^2)}$

6. Originally Posted by xwrathbringerx

The parabola y = x^2 and the line y = mx +b intersect at the points A (α, α^2) and B (β, β^2) as shown.

c) The point P(x, x^2) lies on the parabola between A and B. Show that the area of triangle APB is given by:

Area of Triangle APB = ½ (b + mx – x^2) √ (m^2 + 4b)
AB is the base of triangle, and perpendicular distance from P to AB is height of triangle.

Height = Perpendicular distance from $P(x, x^2)$ to line mx + b - y = 0

$= \frac{|mx + b - x^2|}{\sqrt{1^2+m^2}}$

$= \frac{|mx + b - x^2|}{\sqrt{1+m^2}}$

Area of triangle $= \frac{1}{2}\times base \times height$

$= \frac{1}{2}\times AB \times height$

$=\frac{1}{2}\times \sqrt{(m^2+4b)(1+m^2)} \times \frac{|mx + b - x^2|}{\sqrt{1+m^2}}$

$=\frac{1}{2}(mx + b - x^2) \sqrt{(m^2+4b)}$

$Area,\;\;A(x)=\frac{1}{2}(mx + b - x^2) \sqrt{(m^2+4b)}$

7. Originally Posted by xwrathbringerx

d) Find the position of point P that maximizes the area of triangle APB in terms of m.

d) The area of the triangle is a function with respect to x. Calculate the first derivation and set this term equal to zero

$Area,\;\;A(x)=\frac{1}{2}(mx + b - x^2) \sqrt{(m^2+4b)}$

$A(x)=\dfrac12(mx+b-x^2)\underbrace{\sqrt{m^2+4b}}_{constant!}$

$A'(x)=\dfrac12 \sqrt{m^2+4b} \ \cdot \ (m-2x)$

$A'(x) = 0~\implies~m-2x=0~\implies~x=\dfrac m2$

$
Point \;\;P(x, x^2)=\left(\frac{m}{2}, \frac{m^2}{4}\right)$

Did you get all the parts NOW ???

How exactly does become ?

Is m - 2x the factorised form of mx + b - x^2?

9. Originally Posted by xwrathbringerx
Therefore you only have to derive $(mx+b-x^2)$ with respect to x (the derivation of b wrt x is zero):
$\dfrac{d(mx+b-x^2)}{dx }= m-2x$