Download Basic Calculus - Problem 3 on SOLVING OPTIMIZATION PROBLEMS USING CALCULUS and more Exercises Mathematics in PDF only on Docsity! Basic Calculus Performance Task 2 SOLVING OPTIMIZATION PROBLEMS USING CALCULUS 3. A right circular cylinder is inscribed in a sphere of radius 10 cm. Find the dimensions of the cylinder that has maximum volume. Important Notes: 1. The formula for the volume of a right circular cylinder is V=π x2h, where x is the radius and h is the height. 2. We will earmark x(radius) as rcylinder, h(height) as hcylinder and R (radius) as r sphere for clearer understanding. a. What is the objective? Let it be 𝑉(𝑥), where 𝑥 is the radius of the cylinder. Use 𝑥 as the control variable. The objective is the dimensions of the cylinder. We are required to find the maximum volume of the right circular cylindeer inscribed in a sphere. We let x as the radius of the cylinder, R as the radius of the sphere, and h as the height. b. Sketch the sphere with the inscribed cylinder. Label the important parts. c. Using your illustration, formulate a mathematical expression that represents the height of the cylinder in terms of its radius. From the figure in part b, We can see that we can apply the Pythagorean Theorem to relate the rcylinder (x) in the hcylinder (h). By Pythagorean Theorem, we have: a2+b2=c2→(r¿¿ cylinder)2+( hcylinder2 ) 2 =(r¿¿cylinder )2¿¿→(r¿¿ cylinder)2+( hcylinder2 ) 2 =(10)2 ¿ →(r¿¿ cylinder)2+( hcylinder2 ) 2 =100 ¿ Expressing the hcylinder in terms of its radius: ( hcylinder2 ) 2 =100−(r¿¿cylinder )2¿( hcylinder 2 )=√100−(r¿¿cylinder )2¿ hcylinder=2√100−(r¿¿cylinder)2 ¿ Now, this is the same as h=2√100−x2 [since we let x (the radius of the cylinder) as the control variable] Therefore, the mathematical expression that represents the height of the cylinder in terms of its radius is h=2√100−x2. d. What function accurately models this problem? As stated on the “Important Notes” number 1, we can model the function of this problem by substituting the derived mathematical expression on letter c in the formula V=π x2h. Thus, V=π x2h→=π x2 (2√100−x2 )→2 π x2√100−x2Therefore, the function that accurately models this problem is V ( x )=2π x2√100−x2 e. What are the dimensions of the cylinder? To maximize V ( x ), we will equate its first derivative to zero: V ' ( x )=0 V ' ( x )= d dx (2π x2√100−x2 ) →0=2 π [2x √100−x2+ x2( −2 x 2√100−x2 )] →0=4 π (x √100−x2− x3 2√100−x2 ) →0=x (100−x2 )− x3 2 →0=2 x (100−x2 )−x3
