1
$\begingroup$

How would I solve the following questions?

Sketch the graph of a differentiable function f that satisfies the given conditions if that possible or explain why its not possible:

$f(0)=-3$ $f(3)=0$ $f'(x)<0$

My second question is

Find the greatest possible value of the product $x y$ given that $x$ and $y$ are both positive and $2x+3y=30$

Can anyone please hep me.

$\endgroup$
1
  • $\begingroup$ The function $\frac{2}{x-1}-1$ fulfills the conditions, except for the discontinuity at $x=1$ which is a dealbreaker $\endgroup$
    – DenDenDo
    Commented Mar 19, 2015 at 13:32

5 Answers 5

3
$\begingroup$

(i) We know $f' < 0$, so the function is always decreasing.

(ii)Consider the interval $x\in (-3, 0)$: Note that $f(0) = -3$, and $f(3) = 0$. Hence it seems the function must be increasing on this interval.

Argue that $(i), (ii)$ cannot both possibly true, and therefore, answer: can any function exist with those conditions?


Find the greatest possible value of the product $x y$ given that $x$ and $y$ are both positive and $2x+3y=30$

$$2x + 3y = 30 \implies 3y = 30 - 2x \implies y = 10 - \frac 23 x \tag{1}$$

So we compute, substituting the expression in $(1)$ into $y$: $$ xy = x(10 - \frac 23 x) = 10x - \dfrac 23 x^2 = f(x)\tag{2}$$

Now simplify $(2)$. Then find $f'(x)$ and set $f'(x) = 0$.

If there is a maximum, it will be when $f'(x) = 0$, but you need to test the value where $f'(x) = 0$ to determine whether, in fact, it is a maximum.

Once you determine where $f(x)$ is at it's maximum (the value of $x_{max}$), find $f(x_{max})$: that will be the maximum possible value for $xy$.

$\endgroup$
5
  • $\begingroup$ I got a critical values at x=7.5 $\endgroup$ Commented Mar 7, 2013 at 22:19
  • 1
    $\begingroup$ That will work nicely. Can you find $f(7.5)$? That will be the maximum value of your function $f(x)$, and hence the maximum value of the product for $xy$. $\endgroup$
    – amWhy
    Commented Mar 7, 2013 at 22:22
  • $\begingroup$ yes it does I found that f(7.5)=37.5 is the value of f at the local maximum $\endgroup$ Commented Mar 7, 2013 at 22:25
  • $\begingroup$ That's your answer for your second question! $\endgroup$
    – amWhy
    Commented Mar 7, 2013 at 22:26
  • $\begingroup$ My pleasure, Fernando! $\endgroup$
    – amWhy
    Commented Mar 7, 2013 at 22:26
2
$\begingroup$

For the second one, a somewhat tricky solution: Note that by writing out the squares, $$(2x + 3y)^2 - (2x - 3y)^2 = 24xy.$$ Noting that $2x + 3y = 30$ by definition, we get $$900 - (2x - 3y)^2 = 24 xy.$$ Since we want to maximize $xy$, and the $900$ on the left hand side is constant, maximizing the product $xy$ is equivalent to minimizing $(2x - 3y)^2$. This is obviously minimized when $2x = 3y$, which gives the optimum $x = \frac{15}{2}$, $y = 5$, and $xy = \frac{900}{24} = \frac{75}{2}$.

$\endgroup$
1
$\begingroup$

$f' < 0$ so from left to right the graph slopes up or down?

Now to get from $f(0) = -3$ to $f(3) = 0$ it has to slope which way?

$\endgroup$
3
  • $\begingroup$ The slope is increasing from f(0)-3 to f(3)=0 so therefore it is not possible? $\endgroup$ Commented Mar 7, 2013 at 21:56
  • $\begingroup$ yes indeed they can't both be true $\endgroup$
    – oks
    Commented Mar 7, 2013 at 21:57
  • $\begingroup$ thanks yes that makes sense. $\endgroup$ Commented Mar 7, 2013 at 22:13
1
$\begingroup$

For the first one, $f'(x)<0$ means your function is decreasing, so can those values be taken?

For the second question, do you know Lagrangian multipliers?

$\endgroup$
4
  • $\begingroup$ No I have not heard them yet. $\endgroup$ Commented Mar 7, 2013 at 21:54
  • $\begingroup$ than it will be better to solve for one variable so $2x+3y=30\iff x=\frac{1}{2} ( 30-3y)$ so $x\cdot y= \frac{1}{2} (30-3y) y$ and maximize this one $\endgroup$ Commented Mar 7, 2013 at 21:57
  • $\begingroup$ What does it mean to maximize, is it to find the local maximum? $\endgroup$ Commented Mar 7, 2013 at 22:04
  • $\begingroup$ yes that it does mean, take the derivative and find a zero of it (for $y>0$) $\endgroup$ Commented Mar 7, 2013 at 22:06
1
$\begingroup$

Hint for 1: Mean value theorem

Hint for 2: Solve for one variable in terms of the other, so that $xy$ is now just a function of one variable. If a differentiable function $f(z)$ attains its maximum at $z=t$, then $f'(t)=0$; however, it might be the case that $f'(t)=0$ even without $f(t)$ being a maximum, so you have to do further checking.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .