
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
Step-by-step 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 Â + 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