We're doing optimization problems and this is one that I am having trouble with:
Suppose a business can sell x gadgets for p=250-0.01x dollars apiece, and it costs the business c(x)= 100+25x dollars to produce x gadgets. Determine the production level and cost per gadget required to maximize profit.
I saw in one problem in the book that they determine the production level by setting the derivative of the cost equal to the derivative of the revenue. (I don't know if this is the way that it always needs to be done-- I don't really understand how to find production level.)
Going on this, I know that
p(x)= r(x) - c(x)
where p=profit, r=revenue, and c= cost
which means that p(x)+ c(x) = r(x)
therefore:
250 -.01x + 100 + 25x= 350+24.99x= r(x)
But if we set c'(x)= r'(x) we get
25=24.99 which doesnt work.
Also, if we set p'(x)=0 (to maximize profit) we end up with -.01=0.... which doesn't work, again.
How should I solve this?
To solve this optimization problem, we need to find the production level that maximizes the profit. Let's go through the steps to find the solution.
Step 1: Define the variables:
- x: the number of gadgets produced
- p(x): the price per gadget (revenue)
- c(x): the cost of producing x gadgets
- p(x) - c(x): the profit
Step 2: Write the equations:
Given that the price per gadget is given by p(x) = 250 - 0.01x dollars apiece and the cost of producing x gadgets is c(x) = 100 + 25x dollars, we can calculate the profit function as:
profit = p(x) - c(x)
= (250 - 0.01x) - (100 + 25x)
= 150 - 25.01x
Step 3: Find the derivative of the profit function:
To optimize the profit, we need to find the production level where the derivative of the profit function is zero. Let's find the derivative of the profit function:
profit' = -25.01
Step 4: Set the derivative equal to zero:
To find the production level that maximizes the profit, we set the derivative of the profit function equal to zero:
-25.01 = 0
Step 5: Solve for the production level:
From the equation -25.01 = 0, we see that this equation has no solution. Therefore, the derivative of the profit function does not equal zero, which means there is no maximum profit point.
Step 6: Analyze the limits:
Since we were not able to solve for the production level that maximizes profit using the derivative, we need to analyze the limits. As x approaches infinity, the cost of production (c(x)) increases faster than the revenue (p(x)), resulting in a diminishing marginal profit. As x approaches zero, the cost is constant (100), and the revenue increases (250), resulting in an increasing marginal profit. However, since the cost of production is always positive (100 + 25x), the profit will never be able to exceed the initial revenue of 250. Therefore, the maximum profit occurs when all gadgets are produced and sold.
To find the cost per gadget at this maximum profit point, substitute the production level (x = all gadgets) into the cost function c(x) = 100 + 25x:
cost per gadget = c(x)/x = (100 + 25x)/x
In this case, the cost per gadget would be (100 + 25 * total gadgets)/(total gadgets).
In summary, the production level required to maximize profit is to produce and sell all gadgets. The cost per gadget at this maximum profit point would be equal to (100 + 25 * total gadgets)/(total gadgets).