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Integrative Laptop Case

Decent Essays

The business that I am using will produce protective laptop cases. They cost roughly $30 to produce. The parts needed to produce the product should be kept in consideration, which mainly include a hard and sturdy plastic, molds to shape the cases, and pads for the bottom of the cases. The fixed costs per month are $4000, while the cost function is C(x)=30x+4000.
Once I found the cost function, I found the revenue function. The revenue function is derived by multiplying the price per unit by the number of units. Then I the given price-demand function of x=12000-75p to find p. I solved for p, and found that p=160-x/75. Then I multiplied that equation by x, representing the price per unit. The last function was R(x)=-x^2/75+160x. Once I found the revenue function, I established the feasible range of units demanded per month, which was [0, 12,000] by plugging zero into the original price demand function. After finding the …show more content…

I took the derivative of each functions. The marginal cost function for my business is C’(x)= 30. With a production level of 3000 units per month, I plugged 3000 in for x, however since the marginal cost was just a constant number, it would be 30. Which means that it would cost $30 per unit. The marginal revenue is R’(x)=160- 2/75x, by entering in 3000 in for x creates a marginal revenue which is $80 per unit, leading the revenue to increase to $80 per additional unit added on.
Considering the marginal revenue is larger than the marginal cost, it would increase production which would ultimately increase profit. Then, finally the calculations I made were finding the optimal levels of production. My profit function is P’(x)= 130-(2x/75), so I have to set it equal to 0 then find x, which is 4875. Finally, I plug 4875 into the price function P(x)=(160-x/75) for x, and x is $95. The maximum profit is equals (4875) and the optimal production level should be 4875 laptop cases sold at $95 a

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