Sample Problems, Lesson 4.6
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Example Problem: #10, Lesson 4.6

A box with a square base and open top must have a volume of 32000 cubic centimeters. Find the dimensions of the box that minimize the amount of material used. Given Problem: #10, Lesson 4.6
We make a drawing of the problem. The shading on the top is supposed to remind us that the top is open.
Using the 32000 cubic centimeter volume, we can calculate an expression for the height of the container in terms of x.
We now set up the formula for the surface area of the container, remembering that there is no top on the box. We have the base and 4 sides.

Because we want to find the dimensions that will minimize the surface area, we take the surface area formula and find the derivative.

By setting the derivative equal to zero and solving for x, we can find the x dimension that will minimize the surface area.

We find that x = 40 (and therefore h = 20) are the ideal dimensions.

Example Problem: #18, Lesson 4.6

Find the area of the largest rectangle that can be inscribed in the ellipse: Given Problem: #18, Lesson 4.6
Here we have a sketch of the problem. We are to find the ideal dimensions that will create the inscribed rectangle of the greatest possible area. Using this scheme, the length of this rectangle will be 2x, and the width of this rectangle will be 2y.
Here we find a way to express the variable y in terms of x, so we take the ellipse equation and solve for y.
Now we set up our area formula for the rectangle using our new expression for y and simplifying as much as possible.
Now we take the derivative of the area function.
Now we set the derivative equal to zero and begin simplifying by multiplying both sides of the equation by .
Now we continue to simplify and solve for x. We have now found the best dimension for x.
We now substitute our ideal value for x back into our area function and find the the maximum area for a rectangle inscribed in this ellipse is 2ab.

 

Assignment
1-21, Odds
Lesson 4.6, Pages 312-313

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