# If $a+b=1$ find the greatest value for $a^2b^3$

I have been trying for some time on this question but i am new to inequalities so I am unable to solve it. I tried am gm but failed. Any help would be apriciated

• I presume you want $a$, $b\ge0$? – Lord Shark the Unknown Jan 5 at 6:14
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Assuming that $$a$$ and $$b$$ are nonnegative, apply AM/GM to $$a/2$$, $$a/2$$, $$b/3$$, $$b/3$$ and $$b/3$$ which sum to $$1$$. One gets $$\frac15\ge\sqrt[5]{\frac a2\frac a2\frac b3\frac b3\frac b3}$$ etc.

Just using algebra.

Using $$b=1-a$$, you are looking for the maximum of function $$f(a)=a^2(1-a)^3$$ Compute the derivatives $$f'(a)=-a(1-a)^2 (5 a-2)$$ $$f''(a)=2(1-a)(10 a^2-8 a+1)$$ The first derivative cancels at $$a=0$$, $$a=1$$ and $$a=\frac 25$$. For the first two, $$f(a)=0$$. For the last one $$f\left(\frac{2}{5}\right)=\frac{108}{3125}$$ and $$f''\left(\frac{2}{5}\right)=-\frac{18}{25} <0$$ confirms that this is a maximum value.

• Although it's not the most elegant, this is by far the most straightforward answer! – Toby Mak Jan 5 at 6:55
• You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier. – Toby Mak Jan 5 at 7:02
• Is $f''(x)$ required at all? – Shubham Johri Jan 5 at 7:03
• @ShubhamJohri; Just to confirm that this is a maximum. – Claude Leibovici Jan 5 at 7:08

Does not exist. Try $$b\rightarrow+\infty$$.

For non-negatives $$a$$ and $$b$$ by AM-GM we obtain: $$a^2b^3=2^23^3\cdot\left(\frac{a}{2}\right)^2\left(\frac{b}{3}\right)^3\leq2^23^3\left(\frac{2\cdot\frac{a}{2}+3\cdot\frac{b}{3}}{5}\right)^5=\frac{108}{3125}.$$ The equality occurs for $$\frac{a}{2}=\frac{b}{3},$$ which says that we got a maximal value.

$$f(x)=(1-x)^2\cdot x^3=x^5-2x^4+x^3$$ $$f'(x)=5x^4 -8x^3 + 3x^2=x^2(5x^2-8x+3)=x^2(x-1)(5x-3)$$ $$f'(x) = 0 \implies x \in \{0, 1, \frac 35\}$$ $$f''(x)=20x^3-24x^2+6x=2x(10x^2-12x+3)$$ $$f''(0)=0, f''(1)>0, f''(\frac 35)<0$$

Hence $$b=\frac 35, a=\frac 25$$, if we assume $$a,b > 0$$

• I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same. – Mark Bennet Jan 5 at 6:57
• We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$ – Shubham Johri Jan 5 at 7:06

Let Constraint function be $$C(a,b)= a+b-1=0$$ and the Object function that needs maximization be $$G(a,b)=a^2b^3$$ we have with partial differentiation of combined Lagrangian: $$C(a,b)- \lambda G(a,b)$$ condition to evaluate the multiplier $$\lambda$$ $$\dfrac {\dfrac{\partial C(a,b)}{\partial a} } {\dfrac{\partial C(a,b)}{\partial b} } =\dfrac {\dfrac{\partial G(a,b)}{\partial a}} {\dfrac{\partial G(a,b)}{\partial b} }$$ $$\dfrac{2a^2b^3}{3a^3b^2} =1\quad \rightarrow \dfrac{a}{b} = \dfrac{2}{3}$$

That it attains a maximum.. can be checked with second derivatives of Object function and its sign change.

After plugging in these values Object function maximum value is:

$$\dfrac{27a^5}{8} =\dfrac{4b^5}{9}.$$