I really need help with this Question. The question says ‘ The value of the solid’s surface are is equal to the value of the solid’s volume. Find the value of x.’
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Units of square inches are never equal to units of cubic inches, so we presume the statement of the problem applies to the numerical values in in² and in³.
The volume of a cuboid of length L, width W, and height H is
... V = LWH
The area is
... A = 2(LW + H(L+W))
You want these to be equal for L=9, W=4, and H=x.
Equating the expressions, we have
... 9·4·x = 2(9·4 +x(9+4))
... 36x = 2(36 +13x) . . . . . simplify a bit
... 36x = 72 + 26x . . . . . . simplify more
... 10x = 72 . . . . . . . . . . . . subtract 26x
... x = 7.2
The dimension represented by x is 7.2 inches.
the gift-box above, is pretty much just a rectangular prism, like the one in the picture below, check the picture below.
so as you can see, is really just 6 rectangles stacked up to each other at the edges.
to get the area of the gift-box, we simply get the area of all rectangles.
left and right, is just 2 rectangles of 4*x, or 2(4*x) = 8x.
front and back, is 2 rectangles of 9*x, or 2(9*x) = 18x.
top and bottom, is 2 rectangles of 9*4, or 2(9*4) = 72.
if we sum those up, that's the surface area, SA, of the gift-box, 8x + 18x + 72.
now its volume is simply the product of the height, length and width, namely 9*4*x, or 36x.
[tex]\bf SA=V\implies \stackrel{SA}{8x+18x+72}~~=~~\stackrel{V}{36x}\implies 26x+72=36x \\\\\\ 72=36x-26x\implies 72=10x\implies \cfrac{72}{10}=x\implies \cfrac{36}{5}=x\implies \boxed{7.2=x}[/tex]