Hey everyone this is m@rk and i'm scribing for Wednesday's class. First of we took a quiz on visual calculus about what we've learned from the first and second derivative test. Second , Mr. K introduced a new topic called Optimization Problems. Optimization problems are problems that deal with either finding the maximum or minimum of the function in the problem.
Slide 1
First of all you need to do is draw a diagram to help you visualize the problem. Second, construct the constraint equation. The constraint equation is the equation is something that limits the problem. In this case it is 320=2L+2W. Third, find what you are trying to optimize and construct an optimization equation. In this case you are trying to maximize area, so you need to use the area formula A=LW. Fourth, you need to solve for one of the unknown variables on the constraint equation and then plug it in on the optimization equation. We choose to solve for L in this case and we plug it i! n on the optimization equation. Fifth, solve for the derivative of our new optimization equation. Sixth, solve for the roots of the derivative of our optimization equation. We found out that the root is located at W=80. Seventh, use the first derivative test and find out whether it is a maximum or a minimum. Lastly, answer the question in a complete sentence.
I'm not going to go over the other question since it is just almost the same problem but instead I'm going to give you an outline how to solve an optimization problem, so after digging up the apcalc blog here it is:
Step 1: Find what you are trying to maximize or minimize. This will be stated excplicitly (in the question).
Step 2: Write an equation for it. Use V = for volume and M = for material, A for area, etc. I must insist on using descriptive variables, because in optimization, if you are sloppy, you lose track of what's going on.
Step 3: Try to get the eq! uation into a two variable form, so you can take the derivativ! e.
- Step 3a: To do step 1, you will often have to create a second equation from additional information given in the problem. This may require ingenuity, but it should become natural.
- Step 3b: ISolate one of the two variables in the equation drawn in step 3a.
- Step 3c: In the original equation that you are trying to minimize or maximize, replace the variable you isolated in step 3b.
Step 4: Take the derivative of your two-variable equation.
Step 5: Set the derivative to 0, and solve for the value of the remaining variable.
Step 6: Plus the value of that variable into the first two equations to find all dimensions, including the final goal, such as the amount of material, cost, or volume.
NOTE: This is just an outline so follow it loosely. There is really no chronological way to answer this problems.
Thats it , the next scribe is haiyan.
solve calculus problems
