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.
  • 2
    $\begingroup$ $7\frac 34 $ is the value of $x+y^2$. $\endgroup$ – 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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.