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CAS 14. Suppose we are planning to make a taco from a round tortilla with diameter 8 inches by bending the tortilla so that it is shaped as if it is partially wrapped around a circular cylinder. We will fill the tortilla to the edge (but no more) with meat, cheese, and other ingredients. Our problem is to decide how to curve the tortilla in order to maximize the volume of food it can hold. (a) We start by placing a circular cylinder of radius $r$ along a diameter of the tortilla and folding the tortilla around the cylinder. Let $x$ represent the distance from the center of the tortilla to a point $P$ on the diameter (see the figure). Show that the cross-sectional area of the filled taco in the plane through $P$ perpendicular to the axis of the cylinder is $A(x) = r\sqrt{16 - x^2} - \frac{1}{2}r^2 \sin^{-1}(\frac{2}{r}\sqrt{16 - x^2})$ and write an expression for the volume of the filled taco. (b) Determine (approximately) the value of $r$ that maximizes the volume of the taco. (Use a graphical approach with your CAS.)
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Surface area of a cylinder with fixed volume: $s=\frac{2 \pi r^{3}+2 V}{r}$ It's possible to construct many different cylinders that will hold a specified volume, by changing the radius and height. This is critically important to producers who want to minimize the cost of packing canned goods and marketers who want to present an attractive product. The surface area of the cylinder can be found using the formula shown, where the radius is $r$ and $V=\pi r^{2} h$ is known. Assume the fixed volume is $750 \mathrm{cm}^{3}$. (GRAPH CAN'T COPY) a. Find the equation of the vertical asymptote. How would you describe the nonlinear asymptote? b. If the radius of the cylinder is $2 \mathrm{cm},$ what is its surface area? c. Use a graphing calculator to graph the function on an appropriate window, and use it to find the value of $r$ that minimizes $S(r) .$ That is, find the radius that results in a cylinder with the smallest possible area, while still holding a volume of $750 \mathrm{cm}^{3}$
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Soda can manufacturer wants to minimize the cost of the aluminum used to make their can. The can has to hold volume V of soda. Assuming that the thickness of the can is the same everywhere, the amount of aluminum used to make the can will be proportional to its surface area. That is, suppose the height of the can is h and the radius of the can is r as shown in the figure on the right. A soda can is a circular cylinder with radius r and height h. The curved surface area is 2πrh and the area of each end cap is πr^2. Then the manufacturer wants to minimize $2πrh + 2πr^2h subject to the constraint that πr^2h = V. Here we have used the formulas for the total surface area and volume of a cylinder. Complete parts (a) through (d). (a) A real soda can has volume V = 279 cm^3. By substituting for V in the given equation, write a function of only r: S(r) = (Type an expression: )
Zack A.
The height of the cylinder is 8 inches. We'll be analyzing the surface area of a round cylinder - in other words, the amount of material needed to "make a can". A cylinder (round can) has a circular base and a circular top with vertical sides in between. Let r be the radius of the top of the can and let h be the height. The surface area of the cylinder, A, is A = 2Ï€r^2 + 2Ï€rh (it's two circles for the top and bottom plus a rolled-up rectangle for the side). A round cylinder with a circle top and base with radius r and a height of h. Part a: Assume that the height of your cylinder is 8 inches. Consider A as a function of r, so we can write that as A(r) = 2Ï€r^2 + 16Ï€r. What is the domain of A(r)? In other words, for which values of r is A(r) defined? Part b: Continue to assume that the height of your cylinder is 8 inches. Write the radius r as a function of A. This is the inverse function to A(r), i.e., to turn A as a function of r into r as a function of A. r(A) = Part c: If the surface area is 225 square inches, then what is the radius r? In other words, evaluate r(225). Round your answer to 2 decimal places.
Jacob F.
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