# Optimization (Maxima and Minima)

Question from the book additional mathematics pure and applied by JF Talbert and HH Heng...

Given that $$x + y = 8$$, find the minimum value of $$x + y^2$$.

What I did was as follows;

• I first let $$x + y² = T$$
• I then expressed $$T$$ in terms of y by making $$x = 8 -y$$ from the first equation
• Then I replaced $$y$$ with $$8-y$$ in $$T$$ to get $$8-y+y^2 = T$$
• I then differentiate $$T$$ wrt $$y$$ and I get $$-1+2y$$
• But for the value to be maximum or minimum,the derivative of $$T$$ wrt $$y$$ will be $$= 0$$
• Hence $$0 = -1 + 2y$$
• Making y the subject we get $$y = 1$$
• But earlier we made $$x = 8 -y$$, hence $$x = 8 - \frac{1}{2}$$
• Therefore $$x = 7\frac{1}{2}$$ and $$y = 1$$ But the answer given by the book is $$7\frac{3}{4}$$ but after following the examples, am finding $$\left(7\frac{1}{2},\frac{1}{2}\right)$$ as my answer. Kinda confused on where am going wrong,may anyone please try and attempt it. Your response will be highly appreciated . . . UPDATE: I was only finding the value of y while the question said find the minumum of the whole expression x + y² . The (7.5, .5) I found was to be replaced in the expression in which we're planning to find the minimum value.
• $7\frac 34$ is the value of $x+y^2$. – The Demonix _ Hermit Dec 2 '19 at 12:45

Let T = $$x+y^2$$ where y=8-x $$T=x+(8-x)^2$$ $$T= x^2-15x+64$$ Is upward opening parabola have minimum at vertex at x= 17/2 Min . T will be 31/4