Answer:
Big Container- $0.30
Medium: $0.33
small: $0.25
Step-by-step explanation:
you divided the number of ounces by the cost
for example- 30/9
small would be the best to by because it is the least cost per ounce
A ratio is not multiplication, so I'm just going to throw that out there first in case any other answers trip you up.
9:2 is the correct answer because there are 9 sandwiches for 2 people.
2nd graph (straight line)
<span>Length, width, and height are all 68 cm.
I am assuming that there's a formatting issue with this question and that the actual size limit is 204 cm. With that in mind, let's create a function giving the width of the box in terms of its height. So
w = (204 - h)/2
Now let's create an expression giving the volume of the box in terms of height.
v = lwh
Since the width and length are the same, replace l with w
v = wwh
And now replace w with (102-h/2)
v = (102-h/2)(102-h/2)h
And expand the equation.
v = (102-h/2)(102-h/2)h
v = (10404 -51h - 51h + 0.25h^2)h
v = (10404 -102h + 0.25h^2)h
v = 10404h -102h^2 + 0.25h^3
Since we're looking for a maximum, that can only happen when the slope of the above equation is equal to 0. The first derivative will tell you the slope of the function at each point. So let's calculate the first derivative. For each term, multiply the coefficient by the exponent and then subtract 1 from the exponent. So:
v = 10404h - 102h^2 + 0.25h^3
v = 10404h^1 - 102h^2 + 0.25h^3
v' = 1*10404h^(1-1) - 2*102h^(2-1) +3*0.25h^(3-1)
v' = 10404h^0 - 204h^1 + 0.75h^2
v' = 10404 - 204h^1 + 0.75h^2
We now have a quadratic equation with A=0.75, B=-204, and C=10404. Use the quadratic formula to find the roots, which are 68 and 204. These 2 zeros represent a local minimum and a local maximum. The value 204 is obviously the local minimum since the box would have a width and length of 0 resulting in a volume of 0. So the height must be 68 which means the length and width are (204 - 68)/2 = 136/2 = 68.
To prove that 68 is the optimal height, let's use a height of (68+e) and see what that does to the volume of the box.
v = (102-h/2)(102-h/2)h
v = (102-(68+e)/2)(102-(68+e)/2)(68+e)
v = (102-(34+e/2))(102-(34+e/2))(68+e)
v = (68-e/2)(68-e/2)(68+e)
v = (4624 - 34e - 34 e + 0.25e^2)(68+e)
v = (4624 - 68e + 0.25e^2)(68+e)
v = 314432 - 4624e + 17e^2 + 4624e - 68e^2 + 0.25 e^3
v = 314432 - 51e^2 + 0.25e^3
Now look at the 2 terms that use e. The -51e^2 term will always be negative, but the +0.25e^3 term will be negative if e is negative and positive if e is positive. So a positive e value (e.g. Make the height larger) does have a possibility of increasing the volume if it can overcome the -51e^2 term. So let's make that equation
0 < -51e^2 + 0.25e^3
51e^2 < 0.25e^3
51 < 0.25e
204 < e
So if we make the height 68 + 204 = 272, then we could have a box with a larger volume. But that's impossible since the largest measurement for any edge is 204 and that's assuming you're willing to set the length of the other 2 dimensions to 0.</span>
