Question

HELP WITH CALC! PROBLEM ATTACHED!

Answer 1

here is my thought process in solving this:

since we want to maximize area, that leads me to think I need to derivate an area function.
d( A = xy )
but I need to get y in terms of x.


using similar triangles I can obtain this relationship:
\[\frac {14}{13} = \frac {y}{a} = \frac {b}{x}\]

I only need to know this relationship tho
\[\frac {14}{13} = \frac {y}{a} [Eq 1]\] and
a = 13-x [Eq 2] (is obtained by re arraging 13=x+a)

combine Eq 1 & Eq 2, isolate Y and I get
\[Y= \frac {14}{13} (13-x)\]now y is in terms of x, so I can simplify my area equation from A=xy to
\[A = (x) \frac {14}{13} (13-x)\]

derivate area with respect to x, you get:
\[d( A ) = 14 - \frac{28x}{13}\]set this equal to zero, solve for x.

you now have x, all you need is y.
recall
\[Y= \frac {14}{13} (13-x)\]
plug in the x you got and now you have y.

Answer 2

now i got -6.5 for x and 21 for y...what did i do wrong this time

Answer 3

use x=6.5, not -6.5

Answer 4

is this college calculus or 12th grade?

Answer 5

oh okay i see what i did thank you

Answer 6

^_^
ur welcome
sorry it took so long, its been a long time since i've done calculus