Recent content by pataflora

I have researched the topic. So x = 10/3 will be substituted into the derivative equation, giving a result of 0

By sketching or graphing the original equation to find out that x=10/3 does in fact give a maximum value.

x = 10/3 would give off a maximum volume right? - - - Updated - - - Because the value I get is of about 296 when I substitute x = 10/3 to the original equation.

How do I identify if there is a maximum or a minimum with x = 10/3?

It would not be possible for x = 10 inches because once it is substituted into the original equation it gives a result of 0 - - - Updated - - - So the only x value that we are left with is 10/3

The value of x that makes V’ = 0 is 10 and 10/3 - - - Updated - - - Would that be a minimum or a maximum?

And the derivative of that is V'= 6x^2 - 80x + 200?

The equation in terms of x would be V = 2x^3 - 40x^2 + 200x - - - Updated - - - Right?

You substitute x for H - - - Updated - - - Oh wait do you plug in the equations to each variable to multiply? - - - Updated - - - You are supposed to multiply all those equations?

The question and the answer choices are in the screenshots. I need it asap if possible please. Thanks!