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Calculus: Optimization
Mr. Anderson
Name:____________________________
1. Find two nonnegative numbers whose sum is 90 and whose product is as large as possible. What is
the product?
2. A rectangle is to be inscribed underneath the parabola y  16  x 2 .
largest possible rectangle.
Find the dimensions of the
3. A box is to be made by cutting squares out of the corners of a 10 by 15 inch rectangle sheet. Find
the maximum possible volume.
4.
Find the dimensions of a cylinder with volume 1500 cm3 that use the least amount of material.
Calculus Problem: Optimization
Mr. Anderson
KEY
1. Find two nonnegative numbers whose sum is 90 and whose product is as large as possible. What is
the product?
Let x=first number and y=second number.
X+Y=90 and we want to maximize the function f(X,Y)=XY
If X=90-Y then we can substitute to get a function of y only.
f(Y)=(90-Y)Y=90Y – Y^2
Now, find the critical points. Set the derivative equal to zero, and find Y.
90-2Y=0 becomes Y=45.
Substitute into X=90-Y to find that X=90-45=45
X
0 (endpoint)
45
Y
90
45 (critical point)
Product XY
0
2025 (maximum)
2. A rectangle is to be inscribed underneath the parabola y  16  x 2 .
largest possible rectangle.
Area=Length times Width
Find the dimensions of the
A=LW
Find the x-intercepts.
0=16-X^2
X=4 or X=-4
Therefore the maximum length would be 8 units, but that would leave no height for the rectangle (area would be zero,
clearly not the maximum).
Any other rectangle would have a length: L=2X
Now consider the width. The width will be determined by the y value. Since y=16-X^2, then the width W=16-X^2
Putting these elements together, the area A=LW=(2X)(16-X^2)= 32X-2X^3
To find the critical point, set the derivative equal to zero. Y ‘ = 32 – 6X^2 = 0
Let’s rule out the negative version, since it is the same rectangle.
X
0 (endpoint)
4 (endpoint)
2.3094
W=16-X^2
16
No width
10.667
L=2X
No length
8
4.619
Area
0
0
49.271
Conclusion: The rectangle with the greatest area has length of 4.618 units, a width of 10.667 units, and an area of 49.271
square units.
3. A box is to be made by cutting squares out of the corners of a 10 by 15 inch rectangle sheet. Find
the maximum possible volume.
Volume=(10-2X)(15-2X)X
It is unlikely that you would need to solve this without a calculator. Go ahead and find the maximum using 2 nd calc #5.
Make sure you are using the domain restrictions x: (0,5).
Derivative= 12X^2-100X+150 = 0
You should know, however, that the critical point occurs when the derivative equals zero.
X= 1.962 units
Max Volume: 132 cubic units
4.
Find the dimensions of a cylinder with volume 1500 cm3 that use the least amount of material.
First, let’s assume that the cylinder is made of equally thin material. That means that the less surface area, the less the
material. Volume is what is inside, and it is fixed in this case.
V=1500=(Pi) (R^2) H
and Surface area = 2 (Pi) (R^2) + 2 (Pi) RH
Substitute 1500/(Pi R^2) for H.
Surface area= 2 (Pi) (R^2) + 2 (Pi) R times 1500/(Pi R^2)
Simplify: Surface area= 2 (Pi) (R^2) + 3000* R^-1
Take the derivative and set it equal to zero to find critical points.
4 (Pi) (R) - 3000 R ^-2 =0
Multiply each side by R^2
4 (Pi) (R^3) =3000
R^3 = 3000/ (4*Pi)
Take the cube root.
R=6.2035 cm
Re-substitute to find H=1500/(Pi * 6.2035^2)= 12.407 cm
Conclusion: The design that uses the least amount of material is one with a radius of 6.204 cm and a height of 12.407 cm.
The resulting surface area is 725.395 square cm.