Help with my calculus homework plzzz...?

Question

A coffee manufacturer want to market a new blend of coffee that will cost $3.90 per pound by mixing two coffees that sell for $2.75 and $5 per pound, resptively. What amounts of each coffee should be blended to obtain the desired mixture? Assume that the total weight of the desired blend is 100 pounds....i dont get this one at all please tell me how to get the the answer...i just dont want the answer itself...thanks

Answer 1

This is an algebra question.

Let
x = amount of $2.75/lb coffee
100 - x = amount of $5.00/lb coffee

We have:

2.75x + 5(100 - x) = 3.90*100
2.75x + 500 - 5x = 390
-2.25x = -110
x = -110/-2.25 = 48 8/9

100 - x = 51 1/9

You should use 48 8/9 lbs of $2.75/lb coffee and 51 1/9 lbs of $5.00/lb coffee.

Answer 2

You want to find the amount of $2.75/lb coffee and $5/lb coffee you need to make 100 lbs of $3.90 coffee.
using the info given you cn get the following equations:

a + b = 100 (a is lbs. of $2.75 coffee, b is lbs. of $5/lb coffee. You want a total weight of 100 lbs.)

2.75a + 5b = (3.90)(100) [the price/lbs times number of lbs. gives you the total price of the coffee. You want the combined prices to be equal to the price of 100 lbs. of $3.90/lbs coffee.]

a + b = 100
2.75a + 5b = 390
You need to find 2 variables, and you have two equations. Now you can use the substitution method, linear combination, or graph the two equations and see where they intersect.

Hope that made sense.

Answer 3

tell your teacher that any good coffee manufacturer would want to turn a profit. So the desired mixture would need to cost the company less than $3.90. Any mixture would work as long as it ends up under $3.90

Answer 4

Let he mixes "x" pounds of coffee that cost $2.75. Then the coffee that cost $5 is (100-x). And his selling price is $3.90.

So 2.75x + 5(100-x) = 3.90 *100
or (2.75-5)x = (-5+3.90)*100
or -2.25x = -110
or x = 110/2.25 = 48.89 pound.
or 100-x=51.11 pound
So he has to mix 48.89 pound of the coffee that cost $2.75 per pound, and 51.11 pound of the coffee that cost $5 to sell them at $3.90 per pound.

Answer 5

For the 1st section, take the spinoff and set it equivalent to 0. f(x) = -.005x^2 + x + 5 f'(x) = -.01x+a million -.01x=-a million x=a hundred the optimal top is while the ball has traveled a hundred ft. f(x) = -.005x^2 + x + 5 f(a hundred) = -.0.5*(a hundred)^2 + a hundred + 5 f(a hundred) = 55 the optimal is 55 ft. while it hits the floor, the top would be 0. as a result, f(x) could be 0, as f(x) supplies the top. f(x) = -.005x^2 + x + 5 0 = -.005x^2 + x + 5 nicely, i won't be able to remedy that fairly, so i will take the ordinary/efficient way out and say that the optimal top happens at a million/2 the full horizontal distance. So, if a million/2 the full distance is 55 ft, the full distance (ie: the place the ball hits the floor) is one hundred ten ft. wish that clears issues up.