Maximum/Investigation

[12345786332]

For question 8. I successfully completed the first part, but struggled to do the second, because I'm unsure of the process of finding the maximum of an equation.

I completed the first part of the investigation, but I don't know how to solve parts 2 and especially 3.

 

[tkhunny]

Please show your work. "I successfully completed" is not enough information.

 

[Steven G]

Wow I never knew that you could read off the point B where the max will be.
BTW, this is not meant for the OP.

 

[HallsofIvy]

It would be helpful if you told us what you got for part "a"!
There are, of course, three "similar triangles" in the picture, the entire right triangle with base length 8 cm and height 6 cm, the triangle on top of the rectangle with base length x and height 6- y, and the triangle to the right with base length 8- x and height y.

Since those are similar we must have $\displaystyle \frac{8}{6}= \frac{4}{3}= \frac{x}{6- y}= \frac{8- x}{y}$. From $\displaystyle \frac{4}{3}= \frac{x}{6- y}$, 4(6- y)= 3x so 24- 4y= 3x, 3x+ 4y= 24. From $\displaystyle \frac{4}{3}= \frac{8- x}{y}$, 4y= 3(8- x), 4y= 24- 3x, 3x+ 4y= 24. Those are the same equation which give "y in terms of x" as $\displaystyle y= 6- \frac{3}{4}x$.

The area of the rectangle is $\displaystyle xy= x(6- \frac{3}{4}x)= 6x- \frac{3}{4}x^2= -\frac{3}{4}(x^2- 8x)$. The graph of that is a parabola opening downward. Its maximum value is at the vertex which you can find by "completing the square".

 

[HallsofIvy]

For "2", if p and q are roots of $\displaystyle 2x^2- 5x+ 1= 0$ then we can write $\displaystyle 2x^2- 5x+ 1= 2(x- p)(x- q)= 2(x^2- (p+q)x+ pq)$. You can just "read off" p+q and pq. For $\displaystyle p^2+ q^2$, remember that $\displaystyle (p+ q)^2= p^2+ 2pq+ q^2$. $\displaystyle \frac{1}{p}+ \frac{1}{q}= \frac{q}{pq}+ \frac{p}{pq}= \frac{p+ q}{pq}$.

 

[HallsofIvy]

For "3" if p and q are roots of $\displaystyle 2x^2- 5x+ 1= 2(x- p)(x- q)= 2x^2- 2(p+q)x+ pq= 0$ then "one more than the roots" would be p+ 1 and q+ 1. All quadratic equations having those roots are of the form $\displaystyle a(x- (p+ 1))(x- (q+1))= a(x^2- (p+1)x- (q+1)x+ (p+1)(q+1)= a(x^2- px- x- qx- x+ pq+ p+ q+ 1)= \frac{a}{2}(2x^2- 2(p+q)x+ 2pq)- pq+ p+ q+ 1)$. Since p and q are roots of the original equation, p+ q= 5 and pq= 1.