A painter wants to paint a picture (rectangular portrait) occupying 72 square inches on a canvas allowing a margin of 4 inches on the top and at the bottom and 2 inches on each side. What will be the smallest dimension of the canvas?
12″ x 17″
7″ x 31″
13″ x 16″
10″ x 20″
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CISF-AC-EXE – 2019
– Canvas Width = Picture Width + 2 * Side Margin = $w + 2 \times 2 = w + 4$
– Canvas Height = Picture Height + 2 * Top/Bottom Margin = $h + 2 \times 4 = h + 8$
– The question asks for the smallest dimension of the canvas. Among the given options, we need to find the canvas dimensions $(w+4) \times (h+8)$ such that $w \times h = 72$. We are looking for the pair $(w+4, h+8)$ that represents the smallest overall canvas size (minimum area) or has the smallest minimum dimension among the options.
– The canvas area is $A = (w+4)(h+8) = wh + 8w + 4h + 32$.
– Since $wh = 72$, $A = 72 + 8w + 4h + 32 = 104 + 8w + 4h$.
– To minimize the canvas area, we need to minimize $8w + 4h$ subject to $wh = 72$. This occurs when $8w = 4h$, i.e., $2w = h$.
– Substituting $h = 2w$ into $wh = 72$: $w(2w) = 72 \implies 2w^2 = 72 \implies w^2 = 36$.
– Since dimensions must be positive, $w = 6$. Then $h = 72/6 = 12$.
– Check if $2w=h$: $2 \times 6 = 12$, which is true.
– The picture dimensions that minimize the canvas area are 6 inches by 12 inches.
– The corresponding canvas dimensions are:
– Canvas Width = $w + 4 = 6 + 4 = 10$ inches
– Canvas Height = $h + 8 = 12 + 8 = 20$ inches
– The calculated minimal canvas dimensions are 10″ x 20″, which is option (D).
– Let’s check if the other options correspond to valid picture dimensions and their canvas areas:
– Option A: Canvas 12×17. Picture width = 12-4=8. Picture height = 17-8=9. 8×9=72. Valid. Canvas area = 12×17 = 204.
– Option B: Canvas 7×31. Picture width = 7-4=3. Picture height = 31-8=23. 3×23=69. Not 72. Invalid option.
– Option C: Canvas 13×16. Picture width = 13-4=9. Picture height = 16-8=8. 9×8=72. Valid. Canvas area = 13×16 = 208.
– Option D: Canvas 10×20. Picture width = 10-4=6. Picture height = 20-8=12. 6×12=72. Valid. Canvas area = 10×20 = 200.
– Comparing the valid canvas areas (204, 208, 200), the minimum area is 200 sq inches, corresponding to the 10″ x 20″ canvas. The smallest dimension among the options provided that satisfy the conditions and result in the minimum canvas size is 10 inches (from the 10″x20″ option).