The problem is as follows:

The owner of a bakery invests $3450\,USD$ in order to prepare $150$ cakes of $1\,kg.$ for his shop. Assuming that he would sell each for $(B+2)\,USD$, where $B$ is the maximum integer value which it can take the expression from below:

$$B=(3\cos \theta-5)^2$$

It is known that angle $\theta$ is acute.

Assuming that at the end of the month he is able to sell all his cakes. Find the profit which it was obtained from the sale of all the cakes.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{300 USD}\\ 2.&\textrm{400 USD}\\ 3.&\textrm{350 USD}\\ 4.&\textrm{450 USD}\\ \end{array}$

The thing here I'm assuming is that what it is being requested is in short: Where $P$ is the profit.

$$P=150(B+2)\times 30-3450$$

Thus in the end the question here is how to get the maximum integer value of $B$, as this was specifically stated in the problem.

$B=(3\cos \theta-5)^2$

The only thing which I can remember is that to maximize you could use derivatives. So for that previous equation become into:

$B'=2(3\cos \theta-5)(-3\sin\theta)=-6\sin\theta(3\cos \theta-5)$

From solving this and equating it to zero:

$B=-6\sin\theta(3\cos \theta-5)$

$\sin\theta= 0$

which reduces to

$\theta = 0^{\circ}$


$3\cos \theta-5=0$, but this would mean:


But this solution is not possible in the real set of numbers because $-1<\cos\theta<1$.

In this situation the maximum value attained for that function would be when $\theta = 0^{\circ}$.

This would mean:

$B=(3\cos 0^{\circ}-5)^2=(3-5)^2=4$

But this doesn't seem to be the right approach because the result will at the end be negative?.

Why the approach using derivatives isn't working? Can this be done using another method in precalculus?.

Upon further inspecting the equation. I'm getting the idea that to maximize $B$ I would want $\cos\theta$ to be the less as possible.

Since it mentions that it want an integer value maybe it would mean?

$\cos\theta=\frac{1}{3} \approx 70.53^{\circ}$



Hence $16\times 150$

Another thing which it came to my mind was:

$0<\cos \theta < 1$


$-5 < 3\cos \theta-5<-2$

$25 < (3\cos \theta -5)^2< 4$

Although this doesn't seems logical, but would it meant that

the maximum integer value is 24. Hence


Thus the answer would it be that number?.

Again I'm confused. Can someone help me here?. I'm confused because the sitation doesn't seem to work with derivatives and if using a definition of the boundaries for the cosine function it seems to cause a contradiction. What's wrong here?. Please take the time to read all my effort because it wasn't easy for me and what it would help me the most is an explanation where did my strategy failed?

  • $\begingroup$ if $\theta=90^\circ$, then $B$ would be $25$, but $\theta $ must be less than that $\endgroup$ Jan 8, 2021 at 4:05

2 Answers 2


Since we are told that $\theta$ is acute, we have $0 < \theta < \pi/2$. On this interval, $0 < \cos \theta < 1$, so $3 \cos \theta < 5$. Hence $$B= (3 \cos \theta - 5)^2 = 9 \left(\frac{5}{3} - \cos \theta\right)^2$$ which shows that as $\cos \theta \to \pi/2$, $B$ increases; specifically, at $\theta = 0$, $B = 25$. But since $\theta = 0$ is not attainable, the maximum integer value of $B$ must be $B = 24$, for which we can backsolve and get $\theta = \arccos \frac{5 - 2\sqrt{6}}{3}$. We don't actually need to solve for this specific value of $\theta$, only know that it exists. The profit is therefore $150(26)-3450 = 450.$


Since $\theta$ is acute

$$0 \lt \cos \theta \leq 1$$

$$\implies 0 \lt 3\cos\theta \leq3$$

$$\implies -5 \lt 3\cos\theta -5 \leq 2$$

$$(3\cos\theta-5)^2 \lt 25$$

Therefore $B = 24$

Yes, the final approach that you made is right - there would be some acute theta for which $B=24$, you aren't really concerned in finding the angle, just it's final value in terms of B


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