Maximize Profit from a Quadratic Equation

Question

I've been working with the following word problem:

The following equation represents the number of wooden blocks, $b$, that a company can produce per minute using wooden boards, $w$:

$$b^2 = \frac{w}{2}-2$$

It costs the company \$1 to purchase each board, and the company sells its blocks for \$7 each. How many blocks should the company make per minute make in order to maximize its profit? (Profit is equal to total sales minus total costs.)

I can't seem to find wrap my head around how to solve the problem. The given solution is 2, but I can't seem to arrive at that solution. Any help is appreciated!

Answer 1

First compute the profit function:

$P(b,w) = 7b-w = 7b-2(b^2+2) = -(2b^2-7b+4)$

Maximise that either by setting the derivative to zero or by completing the square. Going for the non-calculus approach,

$P(b) = -2(b-\frac 74)^2 + \frac{17}8$, which reaches a maximum of $\frac{17}8$ when the squared expression is zero, i.e. when $b = \frac 74$

Since the number of blocks must be a whole number to make the problem realistic, test integer values on either side of $\frac 74$, i.e. $1$ and $2$.

You'll find that the profit is higher when $b=2$, and the maximal profit then is $P(2) = 2$.

In fact for other integer values of $b$, you'll find the "profit" is negative, i.e. a loss.

Answer 2

Using calculus approach..

Let $$Profit = Sales - Cost$$ $$= 7b - w$$ $$= 7b - 2(b^{2} + 2)$$

Taking the derivative:

$$P'(b) = 7 - 4b$$

Now set this equal to zero, and solve for the number of blocks:

$$P'(b) = 7 - 4b = 0$$ $$= b = \frac{7}{4}$$

Since partial blocks can't be made, it must be rounded up to 2.