To solve this problem, we need to divide 80 into two parts, say $x$ and $80 - x$, such that the product of the fifth power of one part and the cube of the other is maximized. Let us denote the two parts as:

• $x$: the first part.
• $80 - x$: the second part.

We aim to maximize the function:

$P(x) = x^5 \cdot (80 - x)^3$

Step 1: Differentiate the product function

To find the value of $x$ that maximizes the product, we differentiate $P(x)$ with respect to $x$, and then set the derivative equal to zero.

$P(x) = x^5 \cdot (80 - x)^3$

Using the product rule for derivatives:

$P'(x) = 5x^4 \cdot (80 - x)^3 + x^5 \cdot 3(80 - x)^2 \cdot (-1)$

Simplifying:

$P'(x) = 5x^4(80 - x)^3 - 3x^5(80 - x)^2$

Factor out the common terms $x^4(80 - x)^2$:

$P'(x) = x^4(80 - x)^2 [5(80 - x) - 3x]$

Simplify inside the brackets:

$P'(x) = x^4(80 - x)^2 [400 - 5x - 3x]$ $P'(x) = x^4(80 - x)^2 [400 - 8x]$

Step 2: Set the derivative equal to zero

To maximize the function, we set $P'(x) = 0$:

$x^4(80 - x)^2 (400 - 8x) = 0$

This gives us three possible cases:

1. $x^4 = 0$ implies $x = 0$, but this is not feasible because it would result in no division.
2. $(80 - x)^2 = 0$ implies $x = 80$, but again, this is not feasible since it would mean no division.
3. $400 - 8x = 0$, solving for $x$:

$400 = 8x$ $x = 50$

Thus, one part is $x = 50$, and the other part is $80 - x = 30$.

Step 3: Conclusion

The two parts that maximize the product are 50 and 30. Therefore, the division of 80 that maximizes the product of the fifth power of one part and the cube of the other is $x = 50$ and $80 - x = 30$.

Would you like further details on any part of the solution?

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Related Questions:

1. How do you apply the product rule for derivatives in different scenarios?
2. What role does the critical point play in maximizing a function?
3. Could we solve this using Lagrange multipliers?
4. How does calculus help in optimization problems?
5. What if the numbers were different, how would the approach change?

Tip: Always check the boundary values in optimization problems to ensure you don't miss any maximum or minimum points.