Solve f_t^prime(x)=1/4x*(t-x)+1/8x^2*(-1)

$$f _ { t } ^ { \prime } ( x ) = \frac { 1 } { 4 } x \cdot ( t - x ) + \frac { 1 } { 8 } x ^ { 2 } \cdot ( - 1 )$$

Solve for t

$$\left\{\begin{matrix}\\t=\frac{3x}{2}\text{, }&\text{unconditionally}\\t\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{1}{4}x by t-x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f_{t})x=\frac{1}{4}xt-\frac{1}{4}x^{2}+\frac{1}{8}x^{2}\left(-1\right)$$

Multiply \frac{1}{8} and -1 to get -\frac{1}{8}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f_{t})x=\frac{1}{4}xt-\frac{1}{4}x^{2}-\frac{1}{8}x^{2}$$

Combine -\frac{1}{4}x^{2} and -\frac{1}{8}x^{2} to get -\frac{3}{8}x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f_{t})x=\frac{1}{4}xt-\frac{3}{8}x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{4}xt-\frac{3}{8}x^{2}=\frac{\mathrm{d}}{\mathrm{d}x}(f_{t})x$$

Add \frac{3}{8}x^{2} to both sides.

$$\frac{1}{4}xt=\frac{\mathrm{d}}{\mathrm{d}x}(f_{t})x+\frac{3}{8}x^{2}$$

The equation is in standard form.

$$\frac{x}{4}t=\frac{3x^{2}}{8}$$

Divide both sides by \frac{1}{4}x.

$$\frac{4\times \frac{x}{4}t}{x}=\frac{3x^{2}}{8\times \frac{x}{4}}$$

Dividing by \frac{1}{4}x undoes the multiplication by \frac{1}{4}x.

$$t=\frac{3x^{2}}{8\times \frac{x}{4}}$$

Divide \frac{3x^{2}}{8} by \frac{1}{4}x.

$$t=\frac{3x}{2}$$

Solve for f_t

$$f_{t}\in \mathrm{R}$$

$$x=\frac{2t}{3}\text{ or }x=0$$