Minimizing Area between two functions, one which is constant.

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The discussion focuses on minimizing the area between the curves y = (x+4)(x-5) and y = ax+2 by determining the optimal value of a. The area is calculated using the integral of the difference between the two functions, with ax+2 being the upper function. To find the minimum area, the points of intersection must be established in terms of a, followed by integrating between those limits. The derivative of the area with respect to a is then set to zero to find the minimum. The consensus is that the value of a that minimizes the area is -1, confirming the initial intuition.
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Homework Statement



Graph the curves y = (x+4)(x-5) and y = ax+2 for various values of a. For what value a is the area between the two curves a minimum?

Homework Equations



y = (x+4)(x-5)
y = ax+2

The Attempt at a Solution



For my attempt I attempted to find the primary equation to optimize. I used Area =
\int(ax+2-(x+4)(x-5))
Since when you graph the two ax+2 is the upper function.
I cannot find a secondary equation to bring it to one variable.

The answer itself is a = -1 (guess and check haha) I am just unsure of how to properly go about minimizing this...
a is what is changing in order for the area to be minimized I understand, and I've exhausted all experimentation with finding a secondary equation in my mind.

Any help or input would be greatly appreciated! This problem is already done on my homework (mymathlab so only the answer is needed), I just really would like to figure out how to properly do this!
 
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To solve that problem you have to find the values of x where the two curves intersect in terms of a (use the quadratic equation). Then do the integral between those limits. Finally take the derivative with respect to a and set it equal to zero. It's really quite a mess.
 
Dick said:
To solve that problem you have to find the values of x where the two curves intersect in terms of a (use the quadratic equation). Then do the integral between those limits. Finally take the derivative with respect to a and set it equal to zero. It's really quite a mess.

That's for sure! It's a perfect problem for Maple or its ilk. Maple verifies the OP's intuition that a = -1. Maybe it was a Maple type llab exercise in the first place.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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