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    Break Even Analysis

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    Nature's Green Inc., a manufacturer of alfalfa tablets sold in health - food stores, currently operates just outside of Meno,California. Nature's Green is considering two alternative proposals for expansion, because it has run out of acreage to grow its organically-farmed alfalfa. It has found the following sites where farmers are willing to supply organic alfalfa:

    Alternative 1 : Construct a single plant in Big Cabin, Oklahoma with a monthly production capacity of 50,000 cases, a monthly fixed cost of $275,000, and a variable cost of $100 per case.

    Alternative 2 : Construct three plants, one each in Eudora, Kansas , Springfield,Missouri, and Tonkawa,Oklahoma, with capacities of 25,000 , 20,000 and 15,000 respectively, and monthly fixed costs of $200,000 , $175,000, and $160,000 each. Variable costs would be only $95 per case because of lower distribution costs.

    To achieve these cost savings, sales from each smaller plant would be limited to demand within its home state. The total estimated monthly sales volume of 49,000 cases in these three southeastern states is distributed as follows: 20,000 cases in Kansas , 15,000 cases in Missouri , and 14,000 cases in Oklahoma.

    Assuming a wholesale price of $120 per case, calculate the breakeven output quantities for each alternative
    Assuming sales at the projected levels, which alternative expansion scheme provides Nature's Green with the highest profit per month? If sales increase to production capacities, which alternative would prove to be more profitable?

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    https://brainmass.com/economics/demand-supply/break-even-analysis-289826

    Solution Preview

    Answer 1. BreakEven Quantities

    Alternative 1:
    x*120 = 275,000 + x*100
    => 20x = 275,000
    => x = 13,750

    Alternative 2:
    For Kansas:
    x*120 = 200,000 + x*95
    => 25x = 200,000
    => x = 8,000

    For Missouri:
    x*120 = 175,000 + x*95
    => 25x = 175,000
    => x = 7,000

    For Oklahoma:
    x*120 = 160,000 + x*95
    => 25x = 160,000
    => x = ...

    Solution Summary

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