Maximization problem

Question

The sum of two positive integers is 75. How large can one times the square of the other be?

Answer 1

Maximize y = x²(75 − x).

y' = 150x − 3x².

y'' = 150 − 6x.

Equate y' to 0 or x(150 − 3x) = 0. Then x = 0 or x = 50. Next, y'' = 150 for x = 0 and y'' = -150 for x = 50.

Since y'' is negative for x = 50, it follows that the Maximum of y = x²(75 − x) equals 50²(75 − 50) or 62500 at x = 50.

(Note that y'' being positive for x = 0 indicates a Minimum for y = x²(75 − x) equal to 0 at x = 0.)

Answer 2

Writing out the Equations:

So, I believe you're working with two equations: x+y=75 and x*y²=z.

By using the first equation mentioned above, you can solve for x or y in terms of the other. If I solve for y, for instance, I get y=75-x.

Now, I can substitute y in terms of x into the second equation mentioned above. This gives me x*(75-x)²=z.

Since I want to know the largest value for x*(75-x)², I'll need take the derivative of this equation and set it equal to 0 in order to find it's maximum value between x=0 and x=75.

(Side Note: I know that I am bounded by 0 and 75 because any number greater than 75 or less than 0 will produce a negative integer for either my x value or my y value. I'm only dealing with positive integers in this problem.)

Finding the Derivative:

I can use the power rule and the product rule to take the derivative of x*(75-x)².

This leaves me with x*2*(75-x)*-1+(75-x)²*1=dz/dt. If I simplify this equation, I get

-150x+2x²+5625-150x+x²=dz/dt. Further simplification will give me 1875-100x+x²=dz/dt (combine all of the like terms from the previous equation together, then divide by 3).

Finding the max. and min. values:

Since I want to find the max and min. values of z, I'll need to set dz/dt equal to 0 and solve for x. If I use the quadratic formula, I can convert 1875-100x+x²=0 to (x-25)(x+75)=0. This lets me know that my max and min values from [0,75] are x=25 and x=75.

Now, all I need to do is figure out which of those two x-values is the max and which one is the min. I do that by plugging in both x values into the original equation x*(75-x)²=z. If I plug in x=75, z=0 (min.), but when I plug in x=25, z=62,500 (max.)

Conclusion:

So, therefore, if two integers sum up to 75, one integer times the square of the other can only be as large as 62,500.