The estimated monthly profit (in dollars) realizable by Cannon Precision Instruments for manufacturing and selling x units of its model M1 digital camera is as follows.
p(x)=-0.07x^2+322x-96,000
To maximize its profits, how many cameras should Cannon produce each month?
? cameras
dp/dx = -.14x = 322
= 0 for max of p
.14x = 322
x = 322/.14 = 2300
The weekly demand for the Pulsar 25-in. color console television is given by the demand equation
p = -0.04 x + 598\ \ \ \ \(0<=x<=12,000\)
where p denotes the wholesale unit price in dollars and x denotes the quantity demanded. The weekly total cost function associated with manufacturing these sets is given by
C(x) = 0.000002 x^3 - 0.01 x^2 + 400x + 80,000
where C(x) denotes the total cost incurred in producing x sets. Find the level of production that will yield a maximum profit for the manufacturer. Hint: Use the quadratic formula. (Round your answer to the nearest whole number.)
To maximize profits, we need to find the value of x that corresponds to the maximum value of the profit function p(x)=-0.07x^2+322x-96,000.
To find this value, we can use the vertex formula for a quadratic function: x = -b / (2a)
Here, a = -0.07 and b = 322.
Substituting these values into the formula, we have:
x = -322 / (2*(-0.07))
Simplifying further:
x = -322 / (-0.14)
x = 2300
Therefore, Cannon should produce 2300 cameras each month to maximize its profits.
To maximize profits, we need to find the number of cameras that will yield the maximum value for the profit function p(x). In this case, the profit function is given by:
p(x) = -0.07x^2 + 322x - 96,000
To find the maximum value of p(x), we can use a technique called calculus. We can find the maximum by taking the derivative of p(x) with respect to x, setting it equal to zero, and solving for x. Let's do that.
Step 1: Take the derivative of p(x)
To find the derivative of p(x), we differentiate each term of the function with respect to x:
p'(x) = -0.07(2x) + 322(1) + 0
p'(x) = -0.14x + 322
Step 2: Set p'(x) equal to zero and solve for x
We want to find the value of x when p'(x) equals zero:
-0.14x + 322 = 0
Solving for x:
-0.14x = -322
x = -322 / -0.14
x ≈ 2,300
Step 3: Check the second derivative
To ensure that this value of x corresponds to a maximum point, we need to check the second derivative. Taking the derivative of p'(x) gives us the second derivative:
p''(x) = -0.14
Since the second derivative is negative, it confirms that x ≈ 2,300 is the value that maximizes the profit.
Therefore, Cannon Precision Instruments should produce approximately 2,300 cameras each month to maximize its profits.