2. A cylindrical candle has a volume of 785 cm. Determine the minimum amount of plastic which is needed to cover the outside of the candle for packaging and find the dimensions of the candle which produce this surface area
3. A company which produces pizza ovens has had complaints about their ovens losing too much heat to be
efficient. They have decided to redesign their ovens. The ovens must have a volume of 0.512 m. Find the
optimal design for the oven so that the surface is as small as possible to minimize heat loss Calculate that
surface area.​

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  + 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

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