The question is: What is the maximum value of: $4\sin^2\theta + 3\cos^2\theta$
This is the way I did it:
$4\sin^2\theta + 3\cos^2\theta = \sin^2\theta + 3\sin^2\theta + 3\cos^2\theta = \sin^2\theta + 3$
The max value of $\sin^2\theta$ is $1$, so the answer must be $4$. However my book says that the answer is $5$. Where did I go wrong?
Your argument is OK. You can check it taking the derivative of the function: $$f(x)=4\sin(x)^2+3\cos(x)^2$$ $$y'(x)=2\sin(x)\cos(x)$$ which is null for $x=0,x=\frac{\pi}{2}$ The first value gives $y(x)=3$, with the second one you get $y(x)=4$. So, $4$ is the maximum value of $f(x)$
Though this has been asked and answered, I'll give an alternate solution That avoids messy derivative calculations. Notice that $4\sin^2{x}+3\cos^2{x}$ is very close to $4(\sin^2{x}+\cos^2{x})=4(1)=4$. So we fix it so we get what we want. $$ 4\sin^2{x}+3\cos^2{x}=4\sin^2{x}+3\cos^2{x}+\big(\cos^2{x}-\cos^2{x}\big) $$ Which gives us $$ 4(\sin^2{x}+\cos^2{x})-\cos^2{x}=4-\cos^2{x}=4+(-\cos^2{x}) $$ So the maximum of our original function will occur whenever $-\cos^2{x}$ has a minimum (since we have added it to our original function). Since the minimum of $-\cos^2{x}$ is $0$, the maximum value of our original function $4\sin^2{x}+3\cos^2{x}$ is $4$.
Here's a geometric "solution". With the curve $x(t)= \sqrt{3} \cos t$ and $y(t)=\sqrt{4} \sin t$, $0\leq t \leq 2 \pi$, we get an ellipse with major semiaxis of 2 along the y-axis and minor semiaxis of $\sqrt{3}$ along the x-axis. Then the distance squared to the origin $x(t)^2+y(t)^2=4\sin^2 t + 3 \cos^2 t$ must be maximized to the value of the major semiaxis squared. This is because we can imagine an ellipse inscribed inside a circle to see that it must be 4.
