<PRE><B>Question 1181985</B>
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My visualization of the problem was faulty.  The &quot;pyramids&quot; in the corners of the pan have rectangular bases, not triangular.&lt;br&gt;
Furthermore, the problem is easier than I made it, because the 8x12 dimensions of the bottom of the pan and the 9x13.5 dimensions of the top make those two rectangles similar; and that means the problem can be solved using the idea of a truncated pyramid.&lt;br&gt;
Thanks to tutor @ikleyn for providing a solution by that method.&lt;br&gt;
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UGLY problem....  But I finally decided to buckle down and tackle it.&lt;br&gt;
For calculating the volume of the pan, it can be divided up into several pieces.  For the purpose of this problem, we will express the volume of each piece as a function of h, the height/depth.&lt;br&gt;
(1) The largest piece is a rectangular prism, dimensions (12)x(8)x(h).&lt;br&gt;
Volume of rectangular prism: {{{(12)(8)(h) = 96h}}}.&lt;br&gt;
(2) Next there are triangular prisms along each side of the pan.&lt;br&gt;
The base of the pan is 8x12 inches.  With the sloping sides of the pan, the dimensions of the top edge of the pan are 9x13.5 inches.&lt;br&gt;
That means along the long side of the pan the lengths of the top and bottom edges differ by (9-8)/2 = 0.5 = 1/2 inches; and along the short side of the pan the top and bottom edges differ by (13.5-12)/2 = 0.75 = 3/4 inches.&lt;br&gt;
But those are the &quot;overlaps&quot; for the full pan, with a depth of 2 inches.  For solving this problem we will need to write those overlaps as functions of the depth h.  So the overlap along the long edge, 1/2 inch when the depth is 2 inches, is (1/4)h.  And the overlap along the short edge, 3/4 inch when the depth is 2 inches, is (3/8)h.&lt;br&gt;
So we have two triangular prisms along the two long edges; the legs of the triangular base are h and (1/4)h, and the length is 12.&lt;br&gt;
Volume of those two triangular prisms: {{{(2)((1/2)(h)((1/4)h))(12) = 3h^2}}}&lt;br&gt;
Similarly we have two triangular prisms along the two short edges; the legs of the triangular base are h and (3/8)h, and the length is 8.&lt;br&gt;
Volume of those two triangular prisms: {{{(2)((1/2)(h)((3/8)h))(8) = 3h^2}}}&lt;br&gt;
(3) Lastly, in each of the four corners of the pan we have a pyramid with height h and {{{cross(bases)}}} {{{cross((1/4)h)}}} {{{cross(and)}}} {{{cross((3/8)h)}}} rectangular bases (1/4)h by (3/8)h.&lt;br&gt;
Volume of the four pyramids: {{{cross((4)((1/3)((1/2)((1/4)h)((3/8)h)))(h) = (1/16)h^3)}}} {{{(4)((1/3)(((1/4)h)((3/8)h)))(h) = (1/8)h^3}}}&lt;br&gt;
So the volume of whatever is in the pan as a function of the depth is&lt;br&gt;
{{{V(h) = 96h+6h^2+(1/8)h^3}}}&lt;br&gt;
We are to find the volume of the batter when the pan is filled to half its depth of 2 inches; so we need to evaluate this volume function for h=1.&lt;br&gt;
{{{V(1) = 96+6+1/8}}} = 102.125 cubic inches.&lt;br&gt;
Note the volume of the full pan is V(2) = 192+24+1/2 = 216.5 cubic inches; the amount of batter to fill the pan to half its depth should be a bit less than half the full volume, so our answer of about 102 cubic inches is reasonable.&lt;br&gt;
ANSWER: 102.125 cubic inches of batter will fill the pan to half its depth.&lt;br&gt;
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