oh my goodness.. trigg. problems >_<?

Question

oh my goodness..


The Big Brick Bakery sells more bagels when it reduces its prices, but then its profit changes. The function y = -1000 (x - 0.55) squared + 300 models the bakery's daily profit in dollars from selling bagels, where x is the price of a bagel in dollars. The bakery wants to maximize the profit.


oooyee .. i don't get how to find the maximum profit?? ..confusion >_<

ok .. its not a trigg. problem? woops my bad.. it's a trigg. class course though

Answer 1

It's not a trig problem. Trig problems have sin, cos, tan, cot, sec, or csc.

To find the maximum profit, put the equation in the form of a parabola. The vertex of the parabola represents the maximum profit.

Good luck!

Answer 2

Not trig, calculus.

y = -1000( x - .55 )^2+300

dy/dx ( y ) = -1000*2 ( x - .55 ) * 1 ( by chain rule )

set dx/dy = 0

-2000 ( x - .55 ) = 0, solve for x
x = .55

plug x back into the original function: -1000 ( (.55) - .55)^2+ 300 = 0 + 300 = 300

So you will sell 300 bagels at $ 0.55. Which is the max. This I know because the quadratic is negative.

Answer 3

Multiply out function

= -1000(x-0.55)^2 + 300

= -1000(x^2-1.1x+0.3025) + 300

= -1000x^2 + 1100x - 302.5 + 300

= -1000x^2 + 1100x - 2.5

Take derivative and put = 0

= -2000x + 1100 = 0

2000x = 1100

x = 0.55

So they should sell their bagels for 0.55$

Answer 4

That's not a "trigg" problem. Check your room number, you might be in the wrong room.

Answer 5

Trig, Eh?

- - - This is better solved with CALCULUS.

Take the first derivative, set it to ZERO... solve for a local maximum.

Answer 6

Go to your other question where you asked this. Two of us have answered it now. =P

Use my way though, its easier and makes more sense. ;)