Answer:
a) Revenue function = 1.299r + 1.379p
where: r = gallon of unleaded regular gasoline and p = gallon of regular unleaded premium gasoline
b) cost function = 1.219f + 1.289p
c) profit function = (1.299r + 1.379p) - (1.219f + 1.289p) = 1.299r - 1.219r + 1.379p - 1.289p = 0.08p + 0.09p
d) total profit = (0.08 x 100,000) + (0.09 x 40,000) = $8,000 + $3,600 = $11,600
The area of the triangle is
A = (xy)/2
Also,
sqrt(x^2 + y^2) = 19
We solve this for y.
x^2 + y^2 = 361
y^2 = 361 - x^2
y = sqrt(361 - x^2)
Now we substitute this expression for y in the area equation.
A = (1/2)(x)(sqrt(361 - x^2))
A = (1/2)(x)(361 - x^2)^(1/2)
We take the derivative of A with respect to x.
dA/dx = (1/2)[(x) * d/dx(361 - x^2)^(1/2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(x) * (1/2)(361 - x^2)^(-1/2)(-2x) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2 - x^2 + 361)/(361 - x^2)^(1/2)]
dA/dx = (-2x^2 + 361)/[2(361 - x^2)^(1/2)]
Now we set the derivative equal to zero.
(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0
-2x^2 + 361 = 0
-2x^2 = -361
2x^2 = 361
x^2 = 361/2
x = 19/sqrt(2)
x^2 + y^2 = 361
(19/sqrt(2))^2 + y^2 = 361
361/2 + y^2 = 361
y^2 = 361/2
y = 19/sqrt(2)
We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
I cannot do turning points.
Also, if this is from a test, it can be removed.
But here is a graph of the function.
Answer:
Step-by-step explanation:
Givens
Circle: 3 grams Number: 1 right hand side Number LHS: 2
Square: 2 grams Number: 2 Number LHS: 5
Triangle: x Number: 5 Number LHS: 2
Formula
Mass RHS = Mass LHS
Solution
3*1 + 2*2 + 5x = 3*2 + 5*2 + 2*x Combine
3 + 2 + 5x = 6 + 10 + 2x
5 + 5x = 16 + 2x Subtract 5 from both sides
5-5+5x = 16-5 + 2x Combine
5x = 11 + 2x Subtract 2x from both sides
3x = 11 Divide by 3
x = 11/3
As you can see, I'm not getting any of the answers that you have provided. Ask your teacher how it was done. (The method used is correct).
