The maximum value of the function $y=4{{x}^{2}}+3x+3$ where x varies from 0 to 4?
(a) 118
(b) 79
(c) 3
(d) 10

Hint: Take the term y to the R.H.S and form a quadratic equation in x by assuming that y is a constant. Use the quadratic formula given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve for the value of x in terms of y. Here, a is the coefficient of ${{x}^{2}}$, b is the coefficient of x and c is the constant term. Use the given range of x $0\le x\le 4$ and substitute the value of x obtained above in terms of y. Solve the inequality for the range of y to get the maximum value of y.

Complete step by step solution:
Here we have been provided with the function $y=4{{x}^{2}}+3x+3$ where the value of x ranges from 0 to 4 and we are asked to determine the maximum value of y.
Let us take the term y to the R.H.S, so we get,
$\begin{align}
  & \Rightarrow 0=4{{x}^{2}}+3x+3-y \\
 & \Rightarrow 4{{x}^{2}}+3x+3-y=0 \\
\end{align}$
We can consider the above equation as a quadratic equation in x by considering that y is constant. So using the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, where a is the coefficient of ${{x}^{2}}$, b is the coefficient of x and c is the constant term, to solve for the value of x we get,
$\begin{align}
  & \Rightarrow x=\dfrac{-3\pm \sqrt{{{3}^{2}}-4\left( 4 \right)\left( 3-y \right)}}{2\left( 4 \right)} \\
 & \Rightarrow x=\dfrac{-3\pm \sqrt{9-16\left( 3-y \right)}}{8}..........\left( i \right) \\
\end{align}$
Now, we have been given that x varies from 0 to 4 so mathematically we have $0\le x\le 4$. Substituting the value of x from equation (i) in the given inequality we get,
$\Rightarrow 0\le \dfrac{-3\pm \sqrt{9-16\left( 3-y \right)}}{8}\le 4$
Multiplying all the terms with 8 we get,
$\begin{align}
  & \Rightarrow 0\le -3\pm \sqrt{9-16\left( 3-y \right)}\le 32 \\
 & \Rightarrow 3\le \pm \sqrt{9-16\left( 3-y \right)}\le 35 \\
\end{align}$
On squaring all the terms we get,
\[\begin{align}
  & \Rightarrow 9\le 9-16\left( 3-y \right)\le 1225 \\
 & \Rightarrow 0\le -16\left( 3-y \right)\le 1216 \\
 & \Rightarrow 0\le -48+16y\le 1216 \\
 & \Rightarrow 48\le 16y\le 1264 \\
\end{align}\]
Dividing all the terms with 16 we get,
\[\therefore 3\le y\le 79\]
Therefore, the maximum value of y is 79 in the given range of x. Hence, option (b) is the correct answer.

Note: You can apply some different approaches also to solve the question. One of the approaches is completing the square method. Using this method we will write the given function as $y={{\left( 2x+\dfrac{3}{4} \right)}^{2}}+\dfrac{39}{16}$ where we can clearly see that the maximum of y will be at x = 4. The derivative method can also be applied to get the answer. In the derivative method we will find the point of minima by substituting $\dfrac{dy}{dx}=0$. Here we will get the point $x=\dfrac{-3}{8}$ and we can say that above this value of x the function value will keep on increasing for the increasing values of x. Therefore, for the given condition $0\le x\le 4$ the value of y will be maximum at x = 4.