Economics Math: Constrained optimization problem

Question

I'm so confused by this problem, that I'm not sure if I'm asking the right question, but here goes...

Context
I'm struggling with a constrained optimization problem in economics, where I need to find the maximum utility for given variables.

In the equation below the utility is expressed as a function of the variable t. The y variable is substituted with the constraint 24-t:

U(t,y) = U(t, f(24-t))

I'm told to calculate the derivative of the above equation using the chain rule, then differentiate the production function, and then use the composite function rule. The result is here:

$\frac{dU}{dt} = \frac{∂U}{∂t} - \frac{∂U}{∂y}f'(24-t)$

So far I am fine with this. Now I am told that by equating the derivative to zero, I can maximize the utility. Since $\frac{dU}{dt}=0$, the result looks like this:

$\frac{∂U}{∂t} = \frac{∂U}{∂y}f'(24-t)$

I'm confused by the statement 'by equating the derivative to zero, I can maximize the utility."

Question
Why does equating the derivative to zero, maximize the utility? I hope I'm making sense.

Answer 1

The derivative measures the slope of the curve at the point for which it is being evaluated. Think of the slope of a line tangent to the curve at that evaluation point. When the slope is zero that means that the tangent line is horizontal to the x-axis. Thus the point is either at a peak in the curve or at the bottom of a valley.

Answer 2

Suppose you are simply to choose $x$ to maximize a function $U(x)$.

If $U'(x)>0$ then you could increase $x$ and obtain a higher $U$. Conversely, if $U'(x)<0$ then you could reduce $x$ and that you lead to a higher $U$.

Thus, the $x$ that maximizes $U$ must satisfy $U'(x)=0$.