There is no greatest perimeter of a rectangle with an area of 39 square feet.
To see why, consider the length of 3 and width of 13. Area would be 39 and perimeter would be 3 + 3 + 13 + 13 = 32
Now consider length of 3/2 = 1.5 and 13·2 = 26. Then area is same but perimeter is 1.5 + 1.5 + 26 + 26 = 55.
Now we can just repeatedly half the length and double the width.
L = 3.0, W = 13
⇒ A = 39, P = 32.0
L = 1.5, W = 26
⇒ A = 39, P = 55.0
L = 0.75, W = 52
⇒ A = 39, P = 105.5
L = 0.375, W = 104
⇒ A = 39, P = 208.75
L = 0.1875, W = 208
⇒ A = 39, P = 416.375
L = 0.09375, W = 416
⇒ A = 39, P = 832.1875
L = 0.046875, W = 832
⇒ A = 39, P = 1664.09375
L = 0.0234375, W = 1664
⇒ A = 39, P = 3328.046875
L = 0.01171875, W = 3328
⇒ A = 39, P = 6656.0234375
L = 0.005859375, W = 6656
⇒ A = 39, P = 13312.01171875
Notice how P just kept growing. That's because width grow much faster than length <span>shrinking, </span>so P would grow endlessly.
Hope this helps.
Our three numbers are...
3 3/10 = 3.3
3.1
3 1/4 = 3.25
So, if we order those from least to greatest, we have...
3.1, 3.25, 3.3
which, in the forms given, is...
3.1, 3-1/4, 3-3/10
The adult meal is $38.57 and the student meal is $26.27. You can try it out on a calculator, all you have to do is divide $1350 by 35 and then subtract your answer by 12!
I don't exactly know how to explain step 2 but the working is:
5 - 8x < 2x + 3
5 - 3 < 2x + 8x
2 < 10x
5 < x
(I'm sorry if it's wrong)
Answer:
Step-by-step explanation:
Let
and
be the production level of milk and white chocolate-covered strawberries respectively. According to the given data, we know the total profit will be

The restrictions can be written as



All the restrictions can be plotted in the same graph to find the feasible region where all of them are met. The graph is shown in the image below
The optimal solution will be the level of production such that
* All restrictions are met
* The total profit is maximum
The optimal level of production can be found in (at least) one of the vertices of the feasible region. We'll try each one as follows
P(0,0)=0
P(400,200)=$2.25 (400)+$2.50 (200) = $1400
P(600,200)=$2.25 (600)+$2.50 (200) = $1850
P(800,0)=$2.25 (800)+$2.50 (0) = $1800
We must produce 600 milk chocolate-covered strawberries and 200 white chocolate-covered strawberries to have a maximum profit of $1850/month