optimization of profit and products

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calcissoannoyingugh

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Apr 9, 2021
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someone please help, i just don't know where to start :(, I just need the equations and then I'll know how to solve the rest by myself I hope.
here's the question:

A factory makes empty boxes. They have fixed costs of $100 per day and variable costs of $2 per box. They can sell each box for $(5 − 0.01?) where ? is the number of boxes produced in a day. Find the number of boxes they should produce to maximize daily profit and find that profit.
 
profit = revenue - costs

the whole point of an exercise like this is for you to develop the working equation
 
90% of the practical applications of differential calculus involve finding either (1) the relevant minimum or maximum of the dependent variable of some function or (2) the value or values of the independent variable or variables that maximize or minimize the dependent variable.

So before you can apply the MECHANICS of differential calculus, you must define a function representing a dependent variable determined by one or more independent variable.

For this problem, what is the dependent variable?

How many independent variables do you have?

What is it or what are they?

Can you find a function relating those variables? Only after you do so can you begin the mechanical process of finding derivatives and equating them to zero.

That is pretty much the process in every problem involving optimization of differentiable functions.
 
i mean the problem is I don't really know revenue and profit and all that
but im guessing revenue= x(5- 0.01x) and profit=2x+100
and so when I did all the math i got x=150 boxes and max profit to be $125
would that be right?
 
your expression for revenue is correct, however, 2x+100 is the expression for costs, not profit.

try again ...
 
for some reason i said profit and not costs lol
but i did use that expression (2x+100) for costs
so profit= x(5- 0.01x) - (2x+100)
and after differentiating and all, x=150 and max profit is 125, correct?
 
calcissoannoyingugh said:
for some reason i said profit and not costs lol
but i did use that expression (2x+100) for costs
so profit= x(5- 0.01x) - (2x+100)
and after differentiating and all, x=150 and max profit is 125, correct?
Click to expand...
[MATH]p(x) = x(5 - 0.01x) - (2x + 100) = 3x - 0.01x^2 - 100 \implies [/MATH]
[MATH]p’(x) = 3 - 0.02x \implies [/MATH]
[MATH]p’(x) = 0 \iff x = \dfrac{3}{0.02} = 150 \implies[/MATH]
[MATH]p(150) = 3 * 150 - 0.01 * 150^2 - 100 = 450 - 100 - 0.01 * 22500 = 350 - 225 = 125.[/MATH]
All good. In a more complex problem you might want to evaluate p’’(150) to ensure that you have found a maximum, but here you can see that p(0) = - 100 so p(150) = 125 is certainly not a minimum.
 
calculus is not really necessary to find the value of x that yields maximum profit ...

[MATH]P(x) = x(5-0.01x) - (2x+100) = -0.01x^2 + 3x - 100[/MATH]
parabola has a negative leading coefficient [MATH]\implies[/MATH] it opens downward with vertex at [MATH]x=\frac{-b}{2a} = \frac{-3}{2(-0.01)} = 150[/MATH]
 
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