Answer:
2. Candle dimensions: x = 6.3 cm h = 6.29 cm
A (min) = 373.49 cm²
3. Cylindrical oven dimensions: x = 0.54 m h = 0.55 m
A (min)= 1.4747 m²
Explanation:
2.A The volume V of the cylindrical candle is 785 cm³
V = π*x²*h x is the radius of the base and h the heigh of the cylinder
The surface area A is area of the base π*x² . plus lateral area 2*π*r*h
then . A = π*x² + 2*π*x*h . h = V/π*x²
A as a function of x . is
A(x) = π*x² + 2*π*x*785/π*x²
A(x) = π*x² + 1570/x
Taking derivatives on both sides of the equation we get:
A' (x) = 2*π*x - 1570/x²
A'(x) = 0 . 2*π*x - 1570/x² = 0 . 2*π*x³ = 1570
x³ = 250
x = 6.3 cm . and . h = 785/π*x² . h = 785/124.63
h = 6.29 cm
Then dimensions of the cylindrical candle:
x = 6.3 cm h = 6.29 cm
A (min) = 3.14 * (6.3)² + 6.28*6.3*6.29
A (min) = 124.63 + 248.86
A (min) = 373.49 cm²
3. For a cylindrical oven V = 0.512 h = 0.512/ π*x²
Following the same procedure
A(x) = π*x² + 2*π*x*0.512/π*x² .A(x) = π*x² + 1024/x
A'(x) = 2* π*x - 1.024/x²
A'(x) =0 . 2* π*x - 1.024/x² =0 . 2* π*x³ . = 1.024
x³ = 0.512/π . x³ = 0.163
x = 0.54 m h = 0.512/π*x² . h = 0.55 m
A(min) = 3.14*(0.54)² + 1024/x
A(min)= 0.9156 + 0.5591
A (min)= 1.4747 m²