Calculus I - Optimization

Hide Mobile Notice

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 4.8 : Optimization

Back to Problem List

2. Find two positive numbers whose product is 750 and for which the sum of one and 10 times the other is a minimum.

Show All Steps Hide All Steps

Start Solution

The first step is to write down equations describing this situation.

Let’s call the two numbers $x$ and $y$ and we are told that the product is 750 (this is the constraint for the problem) or,

$xy = 750$

We are then being asked to minimize the sum of one and 10 times the other,

$S = x + 10y$

Note that it really doesn’t worry which is $x$ and which is $y$ in the sum so we simply chose the $y$ to be multiplied by 10.

Show Step 2

We now need to solve the constraint for $x$ or $y$ (and it really doesn’t matter which variable we solve for in this case) and plug this into the product equation.

$x = \frac{{750}}{y}\hspace{0.5in} \Rightarrow \hspace{0.5in}S\left( y \right) = \frac{{750}}{y} + 10y$ Show Step 3

The next step is to determine the critical points for this equation.

$S'\left( y \right) = - \frac{{750}}{{{y^2}}} + 10\hspace{0.5in} \to \hspace{0.5in} - \frac{{750}}{{{y^2}}} + 10 = 0\hspace{0.5in} \to \hspace{0.5in}y = \pm \sqrt {75} = 5\sqrt 3$

Because we are told that $y$ must be positive we can eliminate the negative value and so the only value we really get out of this step is : $y = \sqrt {75} = 5\sqrt 3$ .

Show Step 4

Now for the step many neglect as unnecessary. Just because we got a single value we can’t just assume that this will give a minimum sum. We need to do a quick check to see if it does give a minimum.

As discussed in notes there are several methods for doing this, but in this case we can quickly see that,

$S''\left( y \right) = \frac{{1500}}{{{y^3}}}$

From this we can see that, provided we recall that $y$ is positive, then the second derivative will always be positive. Therefore, $S\left( y \right)$ will always be concave up and so the single critical point from Step 3 that we can use must be a relative minimum and hence must be the value that gives a minimum sum.

Show Step 5

Finally, let’s actually answer the question. We need to give both values. We already have $y$ so we need to determine $x$ and that is easy to do from the constraint.

$x = \frac{{750}}{{5\sqrt 3 }} = 50\sqrt 3$

The final answer is then,

$\require{bbox} \bbox[2pt,border:1px solid black]{{x = 50\sqrt 3 \hspace{0.5in}y = 5\sqrt 3 }}$