Calculus I More Max/Min Problems Practice

This practice exercise covers
Solving Optimization Word Problems

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Make a diagram for questions that need one.

QUESTIONS

1) The sum of two nonnegative numbers is 10. Find these numbers if:

a) their product is a minimum.
b) the sum of their squares is a minimum.
c) the sum of their squares is a maximum.

2) The sum of two nonnegative numbers is 36. Find these numbers if
the first number plus the square of the second is:

a) a maximum.
b) a minimum.

3) A rectangular play yard is being built along the side of a house with fence
on 3 sides and the wall of the house as the 4th side of the yard. If there are
20 metres of fence, find the dimensions that maximize the area of the yard.

4) A farmer with 120 metres of fencing wants to enclose a rectagular pig pen.
The pen will have an internal fence running parallel to one side dividing the area
into two sections, one has twice the length of the other. Find the dimensions to
maximize the area.

5) An 20 cm. square piece of cardboard will be used to create an open box by
cutting equal squares from the corners and folding up the sides.
What dimensions will maximize the volume of such a box?

6) Find the dimensions of the largest rectangle that can be inscribed in a
triangle with sides 6, 8 and 10 centimetres.

7) Find the dimensions of the right triangle with hypotenuse of 2 cm.
with a maximum area.

8) Find the dimensions of the largest rectangle that can be inscribed in a

a) semi-circle with radius 8 cm.
b) a circle with radius 4 cm.

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SOLUTIONS

 1) Let x = 1st part, so 10 – x = 2nd part a) f (x) = x (10 – x) -- a parabola, opens down with a max minimum happens at either endpoint x = 0, or x = 10b) minimize x² + (10 – x)² = f (x) expand to get f (x) = 2x² – 20x + 100 a parabola, opens up with a min so f '(x) = 4x – 20 = 0 minimum at x = 5 and 10 – x = 5.c) maximize x² + (10 – x)² = f (x) since we just found the unique critical pt. which is a min, the max happens at either endpoint x = 0, or x = 10 2) Let x = 1st part, so 36 – x = 2nd part a) f (x) = x + (36 – x)² a parabola that opens up with a minimum maximum happens at the initial endpoint x = 0 since we square the 2nd part, we want it big. first part = 0, so second part = 36b) f (x) = x + (36 – x)² a parabola that opens up with a minimum f '(x) = 2x – 1 = 0 e the minimum happens at x = 0.5 and 36 – x = 35.5. 3) 4) 5) 6) with similar triangles, find y in terms of x. Now So make the base 4 cm and the height 3 cm. 7) Setting the numerator = 0, we get 4b(2 - b²) = 0 and since b can't = 0, the solution is 8) a) Setting the numerator = 0, we get x(32 - x²) = 0 which gives b) Now we have the entire circle with r = 4. and a rectangle above and below the x-axis. so now, A = 4xy instead of 2xy. This time we get x(8 - x²) = 0 . It is a square.

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