VeritasPrepKarishma
You don't need derivatives in CAT either! Anyway, we can solve it either by plugging in numbers or by using this concept:
If the sum of given numbers is constant, their product is maximum when they are equal.
Given: \(2r + (\frac{Q}{2pi})*2*pi*r = 20\)
i.e. \(2r + Qr = 20\)
Maximize \((\frac{Q}{2pi})*pi*r^2 = \frac{Qr^2}{2}\)
Now \(Qr = 20 - 2r\) from above. So, Maximize \((20-2r)r/2 = (10 - r)r\)
Since \(10 - r + r = 10\), to maximize product, \(10 - r = r\) i.e. \(r = 5\)
Responding to a pm:
Quote:
How you come up with this step:
Maximize (Q/2pi)∗pi∗r^2=Qr^2
You state that the maximum comes from the product of the two numbers but wouldn't that product be 2r *(Q/2pi)*2*pi*r = 2Q*r^2?
The question says this: " ... with the greatest possible surface area?"
You need to maximise the surface area of a circular sector. The area of a circle is \(\pi * r^2\).
The area of a circular sector is
\((Q/2\pi) * AreaOfCircle\)
\(= (Q/2\pi) * \pi * r^2\)
To maximise this, we need to maximise \(Qr^2\) basically since the rest are constants.
Using Qr = 20 - 2r, we get that we need to maximise (20 - 2r)*r. Again, this is same as 2*(10 - r)*r so we need to maximise (10 - r)*r (ignoring the constants).
This has a constant sum (10 - r) + r = 10
So product is maximum when (10 - r) = r
Hope this clarifies.