Maximum area of a triangle bounded by the f(x) = (6.3-3x+x^2+.6x^3-.1x^4)

[yourplayin]

I'm trying to find the maximum area of the triangle. My first thought here was to take the derivative of the function. Then set that equal to zero to find the critical numbers. Having the critical numbers would allow me to plug those numbers back in and see which of those numbers gives me the best area of the triangle. This is where I'm stuck though. I'm not sure how to proceed from here.

Critical numbers are x=-1.59318314, x= 0.90786215, and x = 5.18532098.

Would I simply use the area of the triangle formula now? I've been stuck for a few days. Thanks for the help.

 

[Romsek]

Have you plotted f(x)? It's a bimodal curve with x intercepts of x=-3, and x=7.
There is also a local minimum at x=1.

I suspect you'll want to ignore the left part of the curve can work on the maximum triangle bounded by the right part of it.
A vertex will be at the local maxima of the curve and the two legs will descend with a slope as close to the curve as possible.

See if this helps any.

 

[Dr.Peterson]

Before I can think about solving this, I need to know what you mean by "a triangle bounded by the f(x) = (6.3-3x+x^2+.6x^3-.1x^4)".

Does this mean any triangle whose vertices all lie on that curve? Clearly not, since that could be infinite. Perhaps the triangle must lie within the region bounded by the curve and the x-axis? I think you've at least omitted a condition.

Please state the entire problem as given to you, word for word.

 

[yourplayin]

Before I can think about solving this, I need to know what you mean by "a triangle bounded by the f(x) = (6.3-3x+x^2+.6x^3-.1x^4)".

Does this mean any triangle whose vertices all lie on that curve? Clearly not, since that could be infinite. Perhaps the triangle must lie within the region bounded by the curve and the x-axis? I think you've at least omitted a condition.

Please state the entire problem as given to you, word for word.

Find the maximum area of the triangle bounded by??(??)= 6. 3−3x+?^2+ 0.6?^3−0. 1?^4. This is exactly what the question asked.

 

[yourplayin]

Have you plotted f(x)? It's a bimodal curve with x intercepts of x=-3, and x=7.
There is also a local minimum at x=1.

I suspect you'll want to ignore the left part of the curve can work on the maximum triangle bounded by the right part of it.
A vertex will be at the local maxima of the curve and the two legs will descend with a slope as close to the curve as possible.

See if this helps any.

Thank you for the response. I did plot the function in DESMOS. So I do know what the function looks like. I also know that if I plug my critical point of 5.18532098, I get a y value of approx. 28 and some change. But is this the max height? Or do I need you use A=(b*h)/2?