Answer:
The profit will be maximum on x = 250.
Step-by-step explanation:
From the given information:
Revenue = 1500x - x²
Cost = 1500 + 1000x
As we know that
Profit = Revenue - Cost ; Let say it equation 1
Then after putting the values of revenue and cost in equation 1 we have:
Profit = (1500x - x²) - (1500 + 1000x)
Profit = 1500x - x² - 1500 - 1000x
Profit = -x² + 500x - 1500
We know that at the max or min the slope of the graph formed by the profit function will be zero, therefore we find the slope of profit function by taking the first derrivative w.r.t. x as under:
d(Profit)/dx = d/dx(-x² + 500x - 1500)
d(Profit)/dx = -2x + 500
By putting the above slope equal to zero we get:
d(Profit)/dx = -2x + 500 = 0
-2x + 500 = 0
-2x = -500
x = 250
Therefore it is concluded that the profit will be maximum when x will be equal to 250.
The only error you made is on problem 3. Everything else is correct. Nice work.
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Here is how to solve problem 3
Plug x = 1 into the equation and solve for y
-3x + y = 1
-3*1 + y = 1 ... replace x with 1
-3 + y = 1
y - 3 = 1
y - 3 + 3 = 1 + 3 .... add 3 to both sides
y = 4
<h3>The answer is 4</h3>
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Verifying the answer:
Plug (x,y) = (1,4) into the equation. Both sides should be the same number after simplifying both sides.
-3x + y = 1
-3*1 + 4 = 1 ..... replace x with 1; replace y with 4
-3 + 4 = 1
1 = 1
The answer is confirmed.
If you were to graph -3x + y = 1, which is equivalent to y = 3x+1, you'll find that the point (1,4) is on this line.
1)
I:y=3x-4
II:9x-3y=14
substitute y into II:
9x-3*(3x-4)=14
9x-9x+12=14
12=14
this is obviously not equal so there is no solution, the lines are parallel
2)
I:y=4x+6
II:5x-y=6
substitute y into II:
5x-(4x+6)=6
5x-4x-6=6
x=12
substiute x into II:
5*12-y=6
-y=6-60
-y=-54
y=54
the solution is (12,54)
