Revenue=quantity x price= (-2p+1000) (p)= -2p^2+1000p
The maximum revenue will occur when the first derivative is zero so when 2(-2p)+1000=0;p=250
Which generates 125,000 in revenue
Try prices of 245 and 255 and you will see they both are less than 250 thereby proving the max revenue is 250
Answer:
Do you have an attached picture?
Step-by-step explanation:
Answer:
Algebra tiles make solving equations, merging like words, or factoring polynomials a visual operation, similar to students using counters to make sense of addition in elementary school, or students using number lines to make sense of integers.
Step-by-step explanation:
i hope this helped
Answer:
zero goats and 120 Ilamas to get profit of $15,120
Step-by-step explanation:
Goats: G
Ilamas: l
Explicit constraints:
2G + 5l ≤ 400
100G+ 80l≤ 13,200
Implicit constraints
G≥0
I≥0
P= 84G+ 126l
See attachment for optimal area
substituting coordinats of optimal region in profit equation to get profit
When G= 132, l=0
P=84(132) + 126(0)
P=11,088
When G=0, l=120
P=84(0)+ 126(120)
P = 15120
When G= 100, l=40
P=84(100)+126(40)
P=13440
20 cups of 10 oz each it's 200 oz
1 gallon it's 4(32)=128 oz
2 gallons is 256 oz. she will need to buy two gallons
