Find the maximum value of $a^2b^3$ where $a,b$ are positive real numbers satisfying $a+b=10$

Question

Find the greatest value of $a^2b^3$ where $a,b$ are positive real numbers satisfying $a+b=10$.Determine the values of $a,b$ for which the greatest value is attained.

It is my question.I continuously tried to use weighted A.M-G.M. inequality, but unfortunately I found no way.I don't think it can be done using Tchebycheff's inequality or Cauchy-Schwartz's theorem. Please give me any hint for doing that.I also failed to solve another similar problem (but reversed. "Find $\min(3x+2y)$, where $x^2y^3$=48). If possible then please give me hint in that also. Thank you.

Answer 1

By AM-GM inequality, $$\dfrac{\dfrac a2+\dfrac a2+\dfrac b3+\dfrac b3+\dfrac b3}5\ge\sqrt[5]{\dfrac a2\cdot\dfrac a2\cdot\dfrac b3\cdot\dfrac b3\cdot\dfrac b3}$$

Can you use the same idea for $$x+x+x+y+y?$$

Answer 2

About your second problem.

The maximum does not exist.

Try $y\rightarrow0^+$ and $x=\sqrt{\frac{48}{y^3}}$.

If so, $3x+2y\rightarrow+\infty$.

If you mean to find a minimal value of $3x+2y$ then by AM-GM we obtain: $$3x+2y=2\cdot\frac{3x}{2}+3\cdot\frac{2y}{3}\geq5\sqrt[5]{\left(\frac{3x}{2}\right)^2\left(\frac{2y}{3}\right)^3}=10.$$ The equality occurs for $\frac{3x}{2}=\frac{2y}{3}$ and $x^2y^3=48$, which says that $10$ is a minimal value.

Done!

Answer 3

"I don't think it can be done using Tchebycheff's inequality or Cauchy-Schwartz's theorem"

Wait! What kind of pre-calculus course are you taking?

If you know Tchebycheff's inequality or Cauchy-Schwartz's theorem, then you can do some simple caluclus to find max and min.

$f(b) = a^2b^3 = (10-b)^2b^3 = b^5 - 20b^4 + 100b^3$

$f'(b) = 5b^4 - 80b^3 + 300b^2 = 5b^2(b^2 - 16b + 60)$

$f''(b) = 10b(b^2 - 16b + 60) + 5b^2(2b - 16)$

The extrema points are were $f'(b) = 0$.

If $b = 0$ then $f'(0) = 0$ so that is a saddle point. (Obviously $f(0) = 0$ so that can't be a maximum.

$b^2 - 16b + 60 = 0 \implies b = \frac {16 \pm \sqrt{256 - 240}}{2} = 6,10$.

If $b = 10$ then $f''(10) = 100(100 - 160 + 60) + 500(20-16) = 2000 > 0$ so $b = 10$ is a local minimum. (Obviously $f(10) = 0$ so that can't be a maximum.)

If $b =6$ then $f''(6) = 60(36-6*16 + 60) + 5*36(12 - 16) = 0 - 5*36*4 < 0$ so $b= 6$ is the local maximum.

So $a=4; b= 6; a^2b^3 = 4^26^3=3456$ is the maximum.

Answer 4

Use calculus. Simple calculus, differentiate the equation and then equate the differential to zero.