I am not sure how to answer this, it is for a quiz that I am doing and my teacher hasn't specified how to answer this type of problem. I've tried distributing the $x$ out and I got $(x^2-4x+10)/2$ as my minimum area but it's saying I am wrong. I even tried putting an inequality of $>4$ but that's also wrong. I have no idea what to do with this math problem.
Use $A = \frac{1}{2}bh$ to find the function of the area.
$$A = \frac{1}{2}(x)\bigg(\frac{10}{x}+x-4\bigg)$$
$$A = \frac{1}{2}x^2-2x+5$$
Think of this as a function - a quadratic function. The maxima/minima of a function can be found where its first derivative is equal to $0$. This is a quadratic function where the coefficient of the leading term ($a$) is positive. Hence, we can confirm it must be the minimum.
$$A’ = x-2$$
Now, set $A’ = 0$ and see what you get for $x$. Evaluate $A$ (the original function) for that value of $x$ you get.
Edit: If you aren’t familiar with calculus, you could just go with finding the vertex of function $A$. The vertex of the function is at the point $(h, k)$.
$$A = \frac{1}{2}x^2-2x+5$$
$$h = -\frac{b}{2a}$$
$h$ is the $x$-coordinate of the vertex of function $A$. Evaluate it, and just plug in that value in the original function to find $k$, or the $y$-coordinate of the vertex of function $A$. That value of $k$ will be the minimum output, or in this case, the minimum area.
Yet again, you can immediately find $k$ without calculating $h$ first by using a general formula.
$$k = c-\frac{b^2}{4a}$$