Question

Can some one help???? Algebra 1

Answer 1

Lara sells cakes and sandwiches. The daily cost of making cakes is $610 more than the difference between the square of the number of cakes sold and 40 times the number of cakes sold. The daily cost of making sandwiches is modeled by the following equation:

C(x) = 3x2 - 50x + 500

C(x) is the cost in dollars of selling x sandwiches.

Which statement best compares the minimum daily cost of making cakes and sandwiches?

It is greater for sandwiches than cakes because the approximate minimum cost is $210 for cakes and $292 for sandwiches.

It is greater for sandwiches than cakes because the approximate minimum cost is $110 for cakes and $190 for sandwiches.

It is greater for cakes than sandwiches because the approximate minimum cost is $292 for cakes and $210 for sandwiches.

It is greater for cakes than sandwiches because the approximate minimum cost is $190 for cakes and $110 for sandwiches.

Answer 2

Hi there :)

So ummm.. how would you like to solve this?
Since both functions are modeled by parabolas we can put them into vertex form and that way it will be easier to identify the minimums.

Does that approach sound familiar? :o
brb

Answer 3

I think I learned about that

Answer 4

Let's try it with the sandwich function and see if it makes sense.

Answer 5

What is that?

Answer 6

\[\Large\rm C(x) = 3x^2 - 50x + 500\]Factor a 3 out of each x term:\[\Large\rm C(x) = 3\left(x^2 - \frac{50}{3}x\right) + 500\]Err actually, let's write these as decimals, maybe that will make more sense to you.\[\Large\rm C(x) = 3\left(x^2 - 16.7x\right) + 500\]To complete the square we take `half of the b coefficient` and `square it`.
\[\Large\rm (-16.7) \div 2=-8.3\]Then squaring it gives us: \(\Large\rm 69.4\) this is the value that completes the square for us.
So we'll add it inside the brackets.\[\Large\rm C(x) = 3\left(x^2 - 16.7x+69.4\right) + 500\]But we can't simply add inside the brackets like that.
We have to keep things balanced.
So we'll also subtract 69.4 inside the brackets.\[\Large\rm C(x) = 3\left(\color{royalblue}{x^2 - 16.7x+69.4}-69.4\right) + 500\]The blue portion is what we need for our perfect square.
So we'll take the -69.4 out of the brackets.
To do so, we must multiply by 3!\[\Large\rm C(x) = 3\left(\color{royalblue}{x^2 - 16.7x+69.4}\right) -208.3+ 500\]We can write the blue part as a perfect square (x plus half of the b value ):\[\Large\rm C(x) = 3\left(\color{royalblue}{x-8.3}\right)^2 -208.3+ 500\]Oh and let's combine the values on the end,\[\Large\rm C(x) = 3\left(\color{royalblue}{x-8.3}\right)^2 +292\]

Answer 7

So it would be C?

Answer 8

It's a lot of little steps putting it into vertex form.
If you remember vertex form it looks like this:\[\Large\rm y=a(x-h)^2+k\]With a vertex located at (h,k).

So in our case, we only care about the y-coordinate ( that's the one that corresponds to cost).

Answer 9

So it would be C?

Answer 10

?

Answer 11

So here is C:
It is greater for cakes than sandwiches because the approximate minimum cost is
`$292 for cakes`
and
`$210 for sandwiches.`

We were just now working on the `sandwhiches` equation.
What did it give us for a vertex? or just the y value for that matter.

Answer 12

Cause I think you have your values backwards, hmm

Answer 13

Ok so I have the values backwards so it would be 210 for cakes and 292 for sandwiches? Which is A?

Answer 14

@zepdrix

Answer 15

Yah it seems that way -_-
I feel like you're just guessing though lol

Answer 16

No I totally get what you are saying