Honest answer. I forgot to input the equal sign, which forms the following equation: A(area)=[insert equation here]
Also, I thought derivatives traced back all the way to Algebra II. I do not know nothing about derivatives except that I require them to finish this equation.
Thus, my knowledge led me to believe that it was appropriate to post derivatives in the Algebra section.
So, can you help me?
Derivatives are the prime focus of the first course in The Calculus.
If $y=f(x)=kx^n$ then the simple derivative is $\frac{dy}{dx}=f'(x)=n\cdot kn^{n-1}$
If $f(w)=150w^1-\dfrac{70w^2}{40}$ then the derivative is $150(1)w^{1-1}-\dfrac{2(70)w^{2-1}}{40}=150-\dfrac{70w}{20}$
Then, frankly, you shouldn't be taking a course that requires you to take derivatives. You have made it clear that you don't know what derivatives are and that you have difficulty with algebra. And if you do not know that $\displaystyle \frac{70}{20}$ is exactly the same $\displaystyle \frac{7}{2}$ then you need to review arithmetic as well.
The topic of derivatives should not be part of a basic algebra course. That begs a question: Why are you reviewing a topic that is not part of your course? In fact it is a topic that is beyond the scope of your studies. There is something that is really suspicious about this thread. Do you want to tell us what is going on?
Yes. I forgot to mention that Algebra isn't the only thing being covered in our course. But it is my terrible mistake to just dismiss derivatives (in which they are more advanced that Algebra) as algebra material.
Is it possible you have been asked to find the minimum of the given quadratic function, and you did some searching and found that derivatives can be used for that? There are ways to do this specifically for quadratic functions that don't involve differential calculus.
Well damn you. Why did you not say that to begin with?
In other words, you are cheating with a shortcut. You think that you know more that the course designer!
Did it ever occur to you that there may be a very good reason why one may want to find a max/min using algebraic methods?
If you are in an algebra course the use algebra methods. You are CHEATING otherwise. If I find your logon, I will tell your instructor.
I should have said maximum instead of minimum. We see that we have a quadratic whose graph is a parabola opening downwards.
I would begin by reducing the fraction:
$\displaystyle A(w)=150w-\frac{7}{4}w^2$
Now, let's factor:
$\displaystyle A(w)=\frac{1}{4}w(600-7w)$
In this form, it's east to see the roots are:
$\displaystyle w\in\left\{0,\frac{600}{7}\right\}$
Now, the axis of symmetry will be midway between the roots, or
$\displaystyle w=\frac{300}{7}$
And so we can state:
$\displaystyle A_{\max}=A\left(\frac{300}{7}\right)= 150\left(\frac{300}{7}\right)- \frac{7}{4}\left(\frac{300}{7}\right)^2= \frac{2}{7}150^2-\frac{150^2}{7}= \frac{150^2}{7}$
As you mentioned, putting the quadratic into vertex form will work
$\displaystyle A(w)=-\frac{7}{4}w^2+150w=-\frac{7}{4}\left(w^2-\frac{600}{7}+\left(\frac{300}{7}\right)^2\right)+ \frac{7}{4}\left(\frac{300}{7}\right)^2=-\frac{7}{4}\left(w-\frac{300}{7}\right)^2+\frac{150^2}{7}$
This tells us:
$\displaystyle A_{\max}=\frac{150^2}{7}$
And also we know that if given:
$\displaystyle y=ax^2+bx+c$
Then the axis of symmetry lies along the line:
$\displaystyle x=-\frac{b}{2a}$
Applying this to $\displaystyle A(w)$ we find the axis of symmetry is:
$\displaystyle w=-\frac{150}{2\left(-\dfrac{7}{4}\right)}=\frac{300}{7}$
And then we proceed as in the first method I showed above.
Personally, I would wait to study derivatives until you are in a calculus course.
Vertex formula is just too tedious a task to accomplish. Which is why I got creative and researched a new way to do it, and my research lead me to follow the derivative method. I must say it is truly amazing. But I just like using d/dx on quadratics.