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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.
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    $\begingroup$ $7\frac 34 $ is the value of $x+y^2$. $\endgroup$ – The Demonix _ Hermit Dec 2 '19 at 12:45
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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

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