Algebra

\(\text{The expression to be maximized is}\ f(x,y)=xy+y^{3}\).The constraint given is \(x+y=1\), where \(x\) and \(y\) are positive real numbers. The variable \(x\) can be expressed in terms of \(y\) using the constraint: \(x=1-y\). This expression for \(x\) is substituted into the function \(f(x,y)\), resulting in a function of a single variable \(y\): \(f(y)=(1-y)y+y^{3}\). The function is simplified: \(f(y)=y-y^{2}+y^{3}\). To find the maximum value, the derivative of \(f(y)\) with respect to \(y\) is calculated: \(f^{\prime }(y)=\frac{d}{dy}(y-y^{2}+y^{3})=1-2y+3y^{2}\). The critical points are found by setting the derivative equal to zero: \(3y^{2}-2y+1=0\). The discriminant of this quadratic equation is calculated: \(\Delta =b^{2}-4ac=(-2)^{2}-4(3)(1)=4-12=-8\). Since the discriminant is negative (\(\Delta <0\)) and the leading coefficient is positive (\(3>0\)), the quadratic \(3y^{2}-2y+1\) is always positive for all real values of \(y\). This implies that \(f^{\prime }(y)>0\) for all real \(y\), meaning the function \(f(y)\) is strictly increasing. Since \(x\) and \(y\) are positive real numbers and \(x+y=1\), it follows that \(0 <1\). As \(f(y)\) is strictly increasing on the interval \((0,1)\), its maximum value will occur as \(y\) approaches the upper bound of the interval, which is \(y=1\). As \(y\rightarrow 1\), \(x\rightarrow 0\). The value of the function as \(y\) approaches \(1\) is calculated: \(f(1)=(1)(1)+(1)^{3}=1+1=2\). However, since \(x\) and \(y\) must be strictly positive, the maximum value is approached but not strictly attained within the domain. The problem asks for the maximum value, which implies the supremum of the function over the given domain.  Final Answer  The maximum value of \(xy+y^{3}\) is \(2\).\)