Respuesta :
Answer:
Profit function: P(x) = -0.5x^2 + 40x - 300
profit of $50: x = 10 and x = 70
NOT possible to make a profit of $2,500, because maximum profit is $500
Step-by-step explanation:
(Assuming the correct revenue function is 90x−0.5x^2)
The cost function is given by:
[tex]C(x) = 50x + 300[/tex]
And the revenue function is given by:
[tex]R(x) = 90x - 0.5x^2[/tex]
The profit function is given by the revenue minus the cost, so we have:
[tex]P(x) = R(x) - C(x)[/tex]
[tex]P(x) = 90x - 0.5x^2 - 50x - 300[/tex]
[tex]P(x) = -0.5x^2 + 40x - 300[/tex]
To find the points where the profit is $50, we use P(x) = 50 and then find the values of x:
[tex]50 = -0.5x^2 + 40x - 300[/tex]
[tex]-0.5x^2 + 40x - 350 = 0[/tex]
[tex]x^2 - 80x + 700 = 0[/tex]
Using Bhaskara's formula, we have:
[tex]\Delta = b^2 - 4ac = (-80)^2 - 4*700 = 3600[/tex]
[tex]x_1 = (-b + \sqrt{\Delta})/2a = (80 + 60)/2 = 70[/tex]
[tex]x_2 = (-b - \sqrt{\Delta})/2a = (80 - 60)/2 = 10[/tex]
So the values of x that give a profit of $50 are x = 10 and x = 70
To find if it's possible to make a profit of $2,500, we need to find the maximum profit, that is, the maximum of the function P(x).
The maximum value of P(x) is in the vertex. The x-coordinate of the vertex is given by:
[tex]x_v = -b/2a = 80/2 = 40[/tex]
Using this value of x, we can find the maximum profit:
[tex]P(40) = -0.5(40)^2 + 40*40 - 300 = $500[/tex]
The maximum profit is $500, so it is NOT possible to make a profit of $2,500.
Answer:
(A) Profit function is [tex]P(x)=-0.5x^2+40x-300[/tex]
(B) Profit of $50 is at [tex]x=10[/tex] and [tex]x=70[/tex]
(C) The maximum profit is of $500 so, it is impossible to achieve the profit of $2500.
Step-by-step explanation:
Given information:
The function which shows revenue is [tex]90x-0.5x^2[/tex]
The cost function:
[tex]C(x)=50x+300x[/tex]
So,
[tex]R(x)=90x-05x^2[/tex]
As, we know
Profit = Revenue - Cost
[tex]P(x) = 90x-0.5x^2-50x-300\\P(x)=-0.5x^2+40x-300\\[/tex]
On equating with profit of $50,
We get:
[tex]-0.5x^2+40x-300=50[/tex]
[tex]x^2-80x+700=0[/tex]
On solving above equation for the value of x :
We get,
[tex]x_1=70\\x_2=10[/tex]
Hence, for the maximum value of profit the coordinate of vertex is given by:
[tex]x=80/2\\x=40[/tex]
Putting the above value in profit equation
[tex]P(40)=-0.5(40^2)+40(40)-300\\P(40)=500[/tex]
Hence, the maximum profit is of $500 so, it is impossible to achieve the profit of $2500.
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