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.
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