Question 1040207: A manufacturer has found that if his product is priced at N90 per unit then his weekly demand is 50units, but the demand rises to 60units per week if the selling price is N70 per unit. His weekly fixed cost is N3,000 and variable cost N20 per unit. Find the level of production which maximizes profit and determine the maximum profit.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the best that i can figure is that there is a linear relationship between price and quantity that needs to be figured out.
this allows you to express cost in terms of quantity and allows you to express revenue in terms of quantity.
once you have that, you can express profit in terms of revenue and cost.
let's start by finding the relationship between price and quantity.
you are given that, when the price is 90, the quantity is 50.
you are also given that when the price is 70, the quantity is 60.
you have two distinct points that you can form a linear equation from.
the equation will be for price in terms of quantity.
each point is (n,p), where n is the quantity and p is the price.
the first point is (n1,p1) = (50,90).
the second point is (n2,p2) = (60,70).
the linear equation is p = m*n + b, where:
p is the price
n is the quantity
m is the slope
b is the p-intercept which is the value of p when n is equal to 0.
note the similarity between p = mn + b and y = mx + b.
y = mx + b is the same equation as p = mn + b when y is replaced by p and x is replaced by n.
the slope is equal to m which is equal to (p2-p1)/(n2-n1) which is equal to (70 - 90) / (60 - 50) which is equal to -20 / 10 which is equal to -2.
the slope is -2 which is equal to m.
the equation of p = m*n + b becomes p = -2*n + b.
you find the value of b by replacing p and n with the values of one of the points.
we'll use (n1,p1) = (50,90).
when you replace p with 90 and n with 50, the equation of p = -2*n + b becomes 90 = -2*50 + b
solve for b to get b = 90 + 2*50 = 90 + 100 = 190.
if you used (n1,p1) = (60,70), you would have gotten the same value of b.
the equation for p becomes p = -2*n + 190.
this is the price equation.
it solves for price (p) in terms of quantity (n).
it is the linear equation that expresses the relationship between p and n when you are given that (n1,p1) = (50,90) and (n2,p2) = 60,70).
to confirm the equation is correct, replace n with 50 and the equation becomes p = -2*50 + 90 which becomes p = 90.
when you replace n with 60, the equation becomes p = -2*60 + 90 which becomes p = 70.
now you can go back to solve your problem.
the cost equation is c = 3000 + 20n.
c is the cost and n is the quantity.
the revenue equation is r = p*n = (-2n + 190) * n
r is the revenue and p is the price and n is the quantity.
the revenue equation can be simplified to r = -2n^2 + 190n after you perform the multiplication indicated on the right side of the equation.
your profit equation is P = r - c which becomes P = (-2n^2 + 190n) - (3000 + 20n) which can be simplified to P = -2n^2 + 190n - 3000 - 20n which can be further simplified to P = -2n^2 + 170n - 3000.
P is the profit and r is the revenue and c is the cost and n is the quantity.
your profit equation becomes P = -2n^2 + 170n - 3000.
note the use of capital P to distinguish the variable name used for price and the variable name used for profit.
p = small p for price
P = capital p for profit
the profit equation can be simplified to P = -2n^2 + 170n - 3000.
this is a quadratic equation where the general form is P = an^2 + bn + c and therefore a = -2, b = 170, c = -3000
the maximum point on this equation is when n = -b/2a which becomes n = -170 / -4 which becomes 170 / 4 which becomes 42.5.
note that p = -2n^2 + 170n - 3000 is the same equation as y = -2x^2 + 170x - 3000 when you replace p with y and n with x.
note that n = -b/2a is the same equation as x = -b/2x when you replace n with x.
in other words, the formula for the maximum point of the equation of ax^2 + bx + c is x = -b/2a, and likewise, the formula for the maximum point of the equivalent equation of an^2 + bn + c is n = -b/2a.
they are the same equations and follow the same rules, except one equation is expressed in terms of x and the other equation is expressed in terms of n.
y = ax^2 + bx + c when y is the dependent variable and x is the independent variable.
p = an^2 + bn + c when p is the dependent variable and n is the independent variable.
when you graph the price equation, you replace p with y and n with x to get p = -2n + 190 becomes y = -2x + 190.
when you graph the profit equation, you replace P = -2n^2 + 170n - 3000 with y = -2x^2 + 170x - 3000.
this conforms to the requirements of the graphing software that the dependent variable is on the y-axis and the independent variable is on the x-axis.
my graph of the price equation of p = -2n + 190 is shown below:
my graph of the profit equation of P = -2n^2 + 170n - 3000 is shown below:
the quantity that produces the maximum profit is 42.5.
since you can't produce half a unit, then the maximum profit is achievee when the quantity is 42 or 43.
both 42 and 43 will produce the same profit.
when n = 42, P = -2n^2 + 170n - 3000 becomes P = 612
when n = 43, P = -2n^2 + 170n - 3000 becomes P = 612.
since you're dealing with a quadratic equation, the values of P are symmetric about the axis of symmetry which is at n = 42.5, so the results should not be surprising that you have the same values for n = 42 and n = 42 since they're both the same distance fron n = 42.5
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