Optimization with Calculus

You want to make a box to contain dirt and your pet earthworm. Using a 7 in by 10 in rectangle of cardboard, you cut congruent squares from the corners and fold up the sides.

Choose the equation would you use in order to do Calculus to find the maximum volume of dirt (including worm) the box can hold?

• $V=\left(7-2x\right)\left(10-2x\right)$

• $V=x\left(7-2x\right)\left(10-2x\right)$

• $V=x\left(7-x\right)\left(10-x\right)$

• $V=x\left(7+2x\right)\left(10+2x\right)$

What can you conclude from this number line of the derivative of function A(t)?

• there is a maximum for A at t = 12

• there is a minimum for A at t = 12

• there is an inflection point at t = 12

• A is concave up at t = 12

• A is concave down at t = 12

Farmer Jo has 32 square feet of land in which to make an enclosure for bunnies, chicks, and penguins. (see picture)

Choose the equation that represents this information.

• $2x+4y=32$

• $2x+2y=32$

• $xy=32$

• $A=3xy$

same question, but more ...
There will be fencing put in around the entire enclosure and in the middle (as pictured) to create 3 sections, what dimensions should the overall enclosure be in order to use the least amount of fence?

Which equation represents the fence amount (F) that you want to minimize?

•  $F=2x+4y$ 

•  $F=2x+2y$ 

•  $F=xy$ 

•  $F=3\left(x+y\right)$ 

same question, now solve it and answer the question:

Farmer Jo has 32 square feet of land in which to make an enclosure for bunnies, chicks, and penguins.  There will be fencing put in around the entire enclosure and in the middle (as pictured) to create 3 sections, what dimensions should the overall enclosure be in order to use the least amount of fence?

• The overall enclosure should be 8 feet by 8 feet

• The overall enclosure should be 8 feet by 4 feet

• The overall enclosure should be 32 square feet

• The overall enclosure should be  $\sqrt{128}=8\sqrt{2}$  feet on each dimension

Pick the correct interpretation from the given number line for V', the derivative of V.

• V has a minimum at 3

• V has a maximum at both 1 and 5

• V has a maximum at 0 and a minimum at 6

• V has a maximum at 1 and a minimum at 5

 $A'=\frac{-2\left(x-12\right)\left(x+12\right)}{x^2}$  and x has to be positive.
What work makes sense to happen next?

• plug in 12 and -12, resulting in a maximum

• plug in 10 and 14, resulting in a minimum

• plug in 10 and 14, resulting in a maximum

• take the derivative of the expression

A square piece of green origami paper that is 6 inches on a side is being made into a gift box (with no lid) by cutting congruent squares out of each corner, folding up the sides, and taping the edges.

What size squares should you cut out for maximum volume? (do the whole problem)

• I should cut out squares that are 1/2 in by 1/2 in

• I should cut out squares that are 1 in by 1 in

• I should cut out squares that are 3 in by 3 in

• I should not cut out any squares

A closed rectangular shipping box with square base is to be made from 120 square inches of cardboard. What dimensions should the box be for maximum volume?

Choose the constraint and optimization equations that represent the problem. 

•  $x^2=120$  and  $V=x^3$  

•  $2x+y=120$  and   $V=xy$  

•  $x^2y=120$  and  $V=2x^2+4xy$  

•  $2x^2+4xy=120$  and  $V=x^2y$  

A rectangle is bounded by the x-axis and the parabola  $y=12-x^2$ .  What length and width should the rectangle have so that its area is a maximum?  

Given the constraint equation above and the optimization equation  $A=2xy$  , choose the DERIVATIVE of the merged (combined) equation.

•  $A'=2xy$  

•  $A'=\ \ 24-6x^2$  

•  $A'=2x\left(12-x^2\right)$  

•  $A'=12-2x$